A374258 Square array: T(n,k) = ((3^(n+1) + 1)^(k-1) + 2)/3, read by descending antidiagonals.

1, 4, 1, 34, 10, 1, 334, 262, 28, 1, 3334, 7318, 2242, 82, 1, 33334, 204886, 183790, 19846, 244, 1, 333334, 5736790, 15070726, 4842262, 177634, 730, 1, 3333334, 160630102, 1235799478, 1181511766, 129672334, 1595782, 2188, 1, 33333334, 4497642838, 101335557142, 288288870742, 94660803334, 3491569558, 14353282, 6562, 1

Offset: 1

Author

Ahmad J. Masad, Jul 01 2024

Keywords

• nonn
• tabl

Comments

This sequence gives the matrix M in the definition of A365450. Similar to A266577.

Conjecture: For each natural number n, the digits of the product of any (n+1) not necessarily distinct terms of the n-th row in the base (3^(n+1)+1) numeral system appear in nondecreasing order.

Proof of the conjecture. Let b := 3^(n+1)+1. The product of any (n+1) terms of the n-th row has the form p/(b-1), where p is the product of (n+1) numbers of the form b^k+2. Let p = (dm, ..., d1, d0)b, and we have d0+d1+...+dm = 3^(n+1) = b-1. Then p/(b-1) = (dm, ..., d2+...+dm, 1+d1+...+dm)_b, which do form a nondecreasing sequence. - _Max Alekseyev, Jul 03 2024

The preceding result is similar to the property of the nondecreasing products mentioned in A237424. Specifically; the first row of this array is A093137, which is a subsequence of A237424. - Ahmad J. Masad, Jul 30 2024

More generally: Let r and n be positive integers and S be a sequence of all numbers of the form (b^c(1)+b^c(2)+...+b^c(r)+1)/(r+1), where c(1),...,c(r) are nonnegative integers. Then in the numeral system base b := (r+1)^(n+1)+1, the digits of the product of any n+1 (possibly equal) terms of S appear in nondecreasing order. Proof is similar. - Ahmad J. Masad, Jul 30 2024; edited by Max Alekseyev, Aug 01 2024

Examples

			The array begins:
  1    4    34   334   3334
  1   10   262  7318
  1   28  2242
  1   82
  1
Example of the conjecture: Take 5 terms from the 4th row and find their product in base 244 numeral system (since 3^(4+1)+1=244) as follows: 82,19846 twice and 4842262 twice, the product is equal to 82*19846*19846*4842262*4842262 = 757279838666167487626528 = (1, 3, 7, 19, 31, 55, 91, 115, 163, 195, 212)_244 which is in agreement with the conjecture since the digits in 244 base numeral system are in nondecreasing order.
Example of the general property: Take r=3 and n=4, then b=4^5+1=1025. The sequence S is the sequence of the numbers of the form (1025^b(1)+1025^b(2)+1025^b(3)+1)/4. Let's multiply 5 terms of the sequence S, say ((1025^0+1025^0+1025^1+1)/4)*(((1025^0+1025^1+1025^1+1)/4)^2)*(((1025^3+1025^4+1025^4+1)/4)^2) = 257*513^2*552175667969^2 = 20621601208620337073958261562113 = (16,112,308,488,580,680,832,936,964,984,1013)_1025. The digits of the product in base 1025 are in nondecreasing order.
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..820

Crossrefs

Cf. A093137, A266577, A365450, A237424.

Programs

Mathematica

T[n_, k_] := ((3^(n+1) + 1)^(k-1) + 2)/3; Table[T[k, n-k+1], {n, 1, 9}, {k, 1, n}] // Flatten (* Amiram Eldar, Jul 02 2024 *)

PARI

T(n,k) = ((3^(n+1) + 1)^(k-1) + 2)/3 \\ Andrew Howroyd, Jul 01 2024