A157032 Let d(i) be the i-th digit of the decimal expansion of phi=1.6180339887498948482045868...,so that d(0) = 1, d(1) = 6, d(2) = 1, etc. Then a(0) = 1, thereafter a(n) = d(d(n)).

1, 3, 6, 8, 1, 8, 8, 8, 8, 8, 9, 0, 8, 8, 8, 0, 8, 0, 8, 1, 1, 0, 3, 8, 3, 8, 8, 0, 8, 3, 3, 3, 8, 8, 6, 6, 9, 9, 1, 1, 8, 1, 8, 6, 9, 8, 8, 1, 3, 9, 3, 1, 8, 3, 1, 6, 8, 3, 0, 0, 8, 3, 1, 1, 9, 1, 3, 1, 3, 1, 0, 3, 1, 8, 6, 8, 8, 1, 1, 0, 0, 8, 9, 1, 9, 1, 1, 9, 1, 1, 0, 6, 8, 8, 8, 8, 6, 6, 8, 9, 0, 8, 0, 9, 3

Offset: 0

Author

Dan Brown (ddbhockey(AT)hotmail.com), Feb 21 2009

Keywords

• nonn
• base
• easy

Comments

This defines a constant 1.368188888908880808110383880... related to Phi in a peculiar way!

All digits are one of the first 10 digits of Phi = A001622, so 2, 4, 5 and 7 never appear. [From R. J. Mathar, Mar 14 2009]

Crossrefs

Cf. A135725, A155912

Extensions

More digits from R. J. Mathar, Mar 14 2009