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The natural numbers decompose into six categories under the operation of repeated squaring modulo 100, four of which consist of numbers that eventually settle at the attractors 0 (cf. A008592), 1 (cf. A376506), 25 (cf. A017329), or 76 (cf. A376507), and two of which eventually enter one of the 4-cycles 16, 56, 36, 96 (this sequence) or 21, 41, 81, 61 (cf. A376509).

The first-order differences of the numbers in this sequence repeat with a fixed period of length sixteen: 2, 2, 2, 4, 2, 2, 6, 6, 6, 2, 2, 4, 2, 2, 2, 4, ...

The natural numbers decompose into six categories under the operation of repeated squaring modulo 100, four of which consist of numbers that eventually settle at the attractors 0 (cf. A008592), 1 (cf. A376506), 25 (cf. A017329), or 76 (cf. A376507), and two of which eventually enter one of the 4-cycles 16, 56, 36, 96 (cf. A376508) or 21, 41, 81, 61 (this sequence).

The first-order differences of the numbers in this sequence repeat with a fixed period of length sixteen: 6, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 6, 6, ...

This is an old school trick which says that a square of an integer that ends with 5 is easy to compute. Remove the 5, multiply the remaining number by (itself + 1), and concatenate 25 at the end. So, a(n)\100 = A002378(n). - Michel Marcus, Dec 23 2013

A "Goldbach Conjecture" for this sequence: when there are n terms between consecutive odd integers (2n+1) and (2n+3) for n > 0, at least one will be the product of 2 primes (not necessarily distinct). Example: n=3 for consecutive odd integers a(7)=7 and a(11)=9 and of the 3 sequence entries a(8)=12, a(9)=15 and a(10)=16 between them, one is the product of 2 primes a(9)=15=3*5. - Michael Hiebl, Jul 15 2007

A024352 gives distinct values in the array, minus the first row (1, 4, 9, 16, etc.). a(n) gives all solutions to the equation x^2 + xy = n, with y mod 2 = 0, x > 0, y >= 0. - Andrew S. Plewe, Oct 19 2007

Alternatively, triangular sequence of coefficients of Dynkin diagram weights for the Cartan groups C_n: t(n,m) = m*(2*n - m). Row sums are A002412. - Roger L. Bagula, Aug 05 2008

Numbers ending in 2, 3, 5 or 7.

The subsequence of primes is A042993. - Michel Marcus, Jul 19 2015

From Wesley Ivan Hurt, Aug 15 2015, Sep 26 2015: (Start)

Ceiling(a(n)/2) = A047201(n).

Complement of (A197652 Union A262389). (End)

Also, numbers that are congruent to {2, 6, 7, 8, 10} mod 11.

(m^k+1)/11 is a nonnegative integer when

. m is a member of this sequence and k is an odd multiple of 5 (A017329),

. m is a member of A017509 and k is odd but not multiple of 5 (A045572).

If k is even, (m^k+1)/11 is never an integer.

The product of two terms does not belong to the sequence.

Apparently 5, 235 and 72335 are the only terms using digits {2,3,5,7}.

a(n)/5 = {1, 3, 17, 33, 47, 97, 333, 377, 967, 1153, 1517, 3017, 3177, 3333, ...}; terms b(n) that have n 3's must be in the sequence since (5 b(n))^2 yields the decimal number 2 followed by (n-1) 7's then n 2's, and ending in 5 (i.e., 225, 27225, 2772225). Thus 5 b(n) = {15, 165, 1665, 16665, etc.} appears in this sequence. - Michael De Vlieger, Aug 15 2016

All terms are odd multiples of 5 (A017329), i.e., must end in 5, which is the only digit whose square ends in a prime digit. The sequence contains A030487 as an infinite proper subsequence which in turn contains all numbers of the form (5*10^n-5)/3 (these are the above 5 b(n)) as a proper subsequence. - M. F. Hasler, Sep 16 2016

Equally, this is the largest number that can be obtained from the "Choix de Bruxelles", version 1 (A323286) operation applied to n.

Maximal element in row n of irregular triangle in A323460 (or, equally, A323286).

Conjecture: If n contains no digit >= 5, then a(n) = 2*n; otherwise, a(n) is obtained from n by doubling the substring from the last digit >= 5 to the last digit. - Charlie Neder, Jan 19 2019. (This is true. - N. J. A. Sloane, Jan 22 2019)

Corollary: a(n)/n < 10 for all n, and a(n) = 10 - 1/k + O(1/k^2) for n = 10*k+5. - N. J. A. Sloane, Jan 23 2019

The high-water marks for a(n)/n occur at n = 1,15,25,35,45,..., cf. A017329. - N. J. A. Sloane, Jan 23 2019

This number triangle results from the array A(n, m) = T(n+m+1) - T(n-1), with T = A000217, for n, m >= 1. For this array see the example by Bob Selcoe, in A111774 (but with rows continued). The present triangle is obtained by reading the array by upwards antidiagonals: T(n, k) = A(n+1-k, k). See also the Jul 09 2019 comment by Ralf Steiner with the formula c_k(n) (rows k >= 1, columns n >= 3), rewritten for A(n, m) = (m+2)*(2*n+m+1)/2, leading to T(n, k) = (k+2)*(2*n-k+3)/2.

Therefore this triangle is related to the problem of giving the numbers which are sums of at least three consecutive positive integers given as sequence A111774. It allows us to find the multiplicities for the numbers of A111774. They are given in A338428(n).

To obtain the multiplicity for number N (>= 6) from A111774 one has to consider only the triangle rows n = 1, 2, ..., floor((N-3)/3).

The row reversed triangle, considered by Bob Selcoe in A111774, is T(n, n-k+1) = T(n+2) - T(k-1), for n >= 1, and k=1, 2, ..., n.

This triangle contains no odd prime numbers and no exact powers 2^m, for m >= 0. This can be seen by considering the diagonal sequences D(d, k), for d >= 1, k >= 1 or the row sequences of the array A(n, m), for n >= 1 and m >= 1. The result is A(r+1, s-2) = s*(s + 2*r + 1)/2, for r >= 0 and s >= 3 (from the g.f. of the diagonals of T given below). This is also given in the Jul 09 2019 comment by Ralf Steiner in A111774. Therefore A(r+1, s-2) is a product of two numbers >= 2, hence not a prime. And in both cases (i) s/2 integer or (ii) (s + 2*r + 1)/2 integer not both numbers can be powers of 2 by simple parity arguments.

The previous comment means that each T(n, k) has at least one odd prime as a proper divisor.

A number N appears in this triangle, or in A111774, if and only if floor(N/2) - delta(N) >= 1, where delta(N) = A055034(N). For the sequence b(n) := floor(n/2) - delta(n), for n >= 2, see A219839(n), b(1) = -1. See a W. Lang comment in A111774 for the proof.