A113231 Ascending descending base exponent transform of triangular numbers (A000217).

1, 4, 34, 956, 106721, 75818480, 490656737694, 22960404169011552, 7141530219670856270919, 20319415706020976355219258316, 1104797870481014132439711155738991604

Offset: 1

Author

Jonathan Vos Post, Jan 07 2006

Keywords

• easy
• nonn

Comments

A003101 is the ascending descending base exponent transform of natural numbers A000027. The ascending descending base exponent transform applied to the Fibonacci numbers is A113122; applied to the tribonacci numbers is A113153; applied to the Lucas numbers is A113154. Since the parity of the triangular numbers cycles odd, odd, even, even; the parity of this sequence cycles odd, even, even, even. The smallest prime in this sequence is a(5) = 127601. What is the next prime? What is the first triangular value?

Examples

			a(1) = 1 because T(1)^T(1) = 1^1 = 1.
a(2) = 4 because T(1)^T(2) + T(2)^T(1) = 1^3 + 3^1 = 4.
a(3) = 34 = 1^6 + 3^3 + 6^1.
a(4) = 956 = 1^10 + 3^6 + 6^3 + 10^1.
a(5) = 106721 = 1^15 + 3^10 + 6^6 + 10^3 + 15^1.
a(6) = 75818480 = 1^21 + 3^15 + 6^10 + 10^6 + 15^3 + 21^1.
a(7) = 490656737694 = 1^28 + 3^21 + 6^15 + 10^10 + 15^6 + 21^3 + 28^1.
a(8) = 22960404169011552 = 1^36 + 3^28 + 6^21 + 10^15 + 15^10 + 21^6 + 28^3 + 36^1.
a(9) = 7141530219670856270919 = 1^45 + 3^36 + 6^28 + 10^21 + 15^15 + 21^10 + 28^6 + 36^3 + 45^1.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..40

Crossrefs

Cf. A000217, A005408, A113122, A113153, A113154.

Programs

Mathematica

A000217[n_] := Binomial[n + 1, 2]; Table[Sum[A000217[k]^(A000217[n - k + 1]), {k, 1, n}], {n, 1, 10}] (* G. C. Greubel, May 18 2017 *)

PARI

for(n=1,10, print1(sum(k=1,n, (binomial(k+1,2))^(binomial(n-k+2,2))), ", ")) \\ G. C. Greubel, May 18 2017

Formula

a(n) = Sum_{i=1..n} (T(i))^(T(n-i+1)), where T(n) are the triangle numbers.

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

a(n) = Sum_{i=1..n} (A000217(i))^(A000217(n-i+1)).

log(a(n)) ~ n^2 * (-1 + 2*LambertW(2^(-3/2)*exp(1/2)*n))^3 / (8*LambertW(2^(-3/2)*exp(1/2)*n)^2). - Vaclav Kotesovec, Jun 07 2025