A356274 a(n) is the number whose base-(n+1) expansion equals the binary expansion of n.

1, 3, 5, 25, 37, 56, 73, 729, 1001, 1342, 1741, 2366, 2941, 3615, 4369, 83521, 104977, 130340, 160021, 194922, 234741, 280393, 332377, 406250, 474553, 551151, 636637, 732511, 837901, 954304, 1082401, 39135393, 45435425, 52521910, 60466213, 69345326, 79236613

Offset: 1

Author

Thomas Scheuerle, Aug 02 2022

Keywords

• nonn
• base

Comments

If the binary expansion of n is n = bit0*2^0 + bit1*2^1 + bit2*2^2 + ... then a(n) = bit0*(n+1)^0 + bit1*(n+1)^1 + bit2*(n+1)^2 + ... . In other words: Interpret the binary expansion of n as digits in base n+1.

Examples

			n=4 in binary is 100 and interpreting those digits as base n+1 = 5 is a(4) = 25.
		

Crossrefs

Cf. A000523, A007814, A104257, A104258, A128889, A136516.

Programs

Mathematica

a[n_] := FromDigits[IntegerDigits[n, 2], n + 1]; Array[a, 40] (* Amiram Eldar, Aug 19 2022 *)

PARI

a(n) = fromdigits(digits(n, 2), n+1)

Python

def a(n): return sum((n+1)**i*int(bi) for i, bi in enumerate(bin(n)[2:][::-1]))
print([a(n) for n in range(1, 39)]) # Michael S. Branicky, Aug 02 2022

Formula

a(2^n) = (2^n + 1)^n = A136516(n).

a(2^n - 1) = (2^(n^2) - 1)/(2^n - 1) = A128889(n).

a(2^n + 1) = 1 + (2^n + 2)^n.

a(n) = A104257(n+1, n).

a(n) = (1/n)*Sum_{j>=1} floor((n + 2^(j-1))/2^j) * ((n-1)*(n+1)^(j-1) + 1).

a(n) = (1/n)*Sum_{j=1..n} ((n-1)*(n+1)^A007814(j) + 1).

a(2*n) = A104258(2*n+1) - 1.

(2*m+1)^n divides a((2*m+1)^n-1) for positive m and n > 0.