Page 57 is not encrypted and can be decoded directly from the Gematria Primus:

Parable : like the instar tunneling to the surface. We must shed our own circumferences. Find the divinity within and emerge.

This is Parable 1,595,277,641 which appeared in 2013 within the ID tag of 761.mp3. The mp3 title is 'The Instar Emergence' which can be summed using the Gematria Primus to obtain '761'. The number 1,595,277,641 is similarly obtained using the product of the gematria sums of each line of the parable ie 1259*1031*1229 = 1,595,277,641.

## Symbology[edit | edit source]

The Mayfly is an alteration of pages 23 to 26 mayflies.

Its original source can be found here : http://www.metafysica.nl/wings/wings_14.html, Figure 4, Mayfly (5).

## Cryptography[edit | edit source]

## Steganography[edit | edit source]

## Numerology[edit | edit source]

## Loose Ends[edit | edit source]

Interesting observation speaking of circumferences:

Way back on Onion 1, the Blake Collage.

http://uncovering-cicada.wikia.com/wiki/DEBATE_ABOUT_MEANING_OF_THE_PICTURE_ON_ONION

It was thought that 1033 was written in roman numerals at the end of the pointing finger.

1 + 0 + 3 + 3 = 7, so if the center point of the circle is 7, lets say we can assume the diameter is 14. The circumference would be 43.982... If we drop all digits after the decimal (because Babylonians didn't use decimals) then we get 43.

43 is prime and the 14th prime. The sum of the digits is 7.

If you spell 43 as Fourty three, rather than Forty three, then two Gematria values are possible because of the TH in three.

(T & H) = 2 + 7 + 3 + 11 + 59 + 103 + (59 + 23) + 11 + 67 + 67 = 412 = 4 + 1 + 2 = 7

(TH) = 2 + 7 + 3 + 11 + 59 + 103 + (5) + 11 + 67 + 67 = 335 = 3 + 3 + 5 = 11 prime

The funny thing is 11 is called A Pointer prime and the finger in the painting is pointing at 1033. :)

Using pixel measurements I calculated that the image is 188 pixels high. As that is the longer side it would also be the diameter if you drew a circle perfectly around the image its circumference would be C=D*pi If D=188 The 188*pi=509.169 so the circumference of a circle around the image would be 509.169 pixels. Not sure if this helps.