cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A286947 Triangle read by rows in which row(n) = {T(n, k)} is the lexicographically earliest list of n numbers such that adding 1 to some T(n, k) gives a row of numbers each divisible by prime(k).

Original entry on oeis.org

1, 3, 2, 15, 20, 24, 105, 140, 84, 90, 1155, 770, 924, 1980, 2100, 15015, 10010, 24024, 4290, 13650, 23100, 255255, 340340, 204204, 364650, 464100, 353430, 60060, 4849845, 6466460, 5819814, 1385670, 3527160, 5969040, 570570, 510510, 111546435, 74364290, 44618574, 127481640, 81124680, 102965940, 39369330, 58708650, 29099070
Offset: 1

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Author

David A. Corneth, May 17 2017

Keywords

Comments

1 + the Rowsum of row(n) gives a multiple of A002110(n).
c = Product_{i=1..n} prime(i)^T(n, i) is the least term such that prime(i) * c is a prime(i)-th power. First such terms are 2, 72, 6810125783203125000000000000000, ... which relates this sequence to A286930.
T(n,k) is a multiple of A258566(n,k). - Peter Munn, Jan 13 2018

Examples

			Row(1): [1]
Row(2): [3, 2]
Row(3): [15, 20, 24]
Row(4): [105, 140, 84, 90]
Row(5): [1155, 770, 924, 1980, 2100]
Row(6): [15015, 10010, 24024, 4290, 13650, 23100]
Row(7): [255255, 340340, 204204, 364650, 464100, 353430, 60060]
Row(8): [4849845, 6466460, 5819814, 1385670, 3527160, 5969040, 570570, 510510]
Row(4) = [105, 140, 84, 90].
Adding 1 to T(4, 1) gives [106,140,84,90], all elements divisible by prime(1) = 2.
Adding 1 to T(4, 2) gives [105,141,84,90], all elements divisible by prime(2) = 3.
Adding 1 to T(4, 3) gives [105,140,85,90], all elements divisible by prime(3) = 5.
Adding 1 to T(4, 4) gives [105,140,84,91], all elements divisible by prime(4) = 7.
The sum of elements in row 3 is 15 + 20 + 24 = 59. 59 + 1 = 60, a multiple of A002110(3) = 30.
		

Crossrefs

Programs

  • PARI
    row(n) = my(pr=primes(n), p = prod(i=1, #pr, pr[i]), res=vector(n, i, lift(chinese(Mod(-1, pr[i]), Mod(0, p/pr[i]))))); res

Formula

T(n, 1) = A002110(n) / 2.
For n >= 2, T(n,n) = A075306(n-1) - 1. - Peter Munn, Jan 13 2018

Extensions

Name corrected by Peter Munn, Jan 12 2018