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

1,2

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.

Links

Table of n, a(n) for n=1..37.

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.

Example

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

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A290509 A347947 A208800 * A249935 A364961 A009002

Adjacent sequences: A356271 A356272 A356273 * A356275 A356276 A356277

Keyword

nonn,base

Author

Thomas Scheuerle, Aug 02 2022

Status

approved