cp's OEIS Frontend

Row sums of the Euler related triangle A117411. Partial sums are A006495. - Paul Barry, Mar 16 2006

Binomial transform of [1, 0, -4, 0, 16, 0, -64, 0, 256, 0, ...], i.e. powers of -4 with interpolated zeros. - Philippe Deléham, Dec 02 2008

The absolute values of these numbers are the odd numbers y such that x^2 + y^2 = 5^n with x and y coprime. See A098122. - T. D. Noe, Apr 14 2011

Pisano period lengths: 1, 1, 8, 1, 4, 8, 48, 4, 24, 4, 60, 8, 12, 48, 8, 8, 16, 24, 90, 4, ... - R. J. Mathar, Aug 10 2012

Multiplied by a signed sequence of 2's we obtain 2, -2, -6, 22, -14, -82, 234, -58, -1054, 2398, 474, -12938, ..., the Lucas V(-2,5) sequence. - R. J. Mathar, Jan 08 2013

Equals the Hausdorff dimension of the Sierpinski fractal square-based pyramid, when each square-based pyramid is replaced by 5 half-size such square-based pyramids (see IREM link). - Bernard Schott, Nov 30 2022

a(n)=1 iff n=A007694(k) for some k.

Numerator of phi(n)/n=Prod_{p|n} (1-1/p). - Franz Vrabec, Aug 26 2005

From Wolfdieter Lang, May 12 2011: (Start)

For n>=2, a(n)/A109395(n) = sum(((-1)^r)*sigma_r,r=0..M(n)) with the elementary symmetric functions (polynomials) sigma_r of the indeterminates {1/p_1,...,1/p_M(n)} if n = prod((p_j)^e(j),j=1..M(n)) where M(n)=A001221(n) and sigma_0=1.

This follows by expanding the above given product for phi(n)/n.

The n-th member of this rational sequence 1/2, 2/3, 1/2, 4/5, 1/3, 6/7, 1/2, 2/3, 2/5,... is also (2/n^2)*sum(k,with 1<=k=2.

Therefore, this scaled sum depends only on the distinct prime factors of n.

See also A023896. Proof via PIE (principle of inclusion and exclusion). (End)

In the sequence of rationals r(n)=eulerphi(n)/n: 1, 1/2, 2/3, 1/2, 4/5, 1/3, 6/7, 1/2, 2/3, 2/5, 10/11, 1/3, ... one can observe that new values are obtained for squarefree indices (A005117); while for a nonsquarefree number n (A013929), r(n) = r(A007947(n)), where A007947(n) is the squarefree kernel of n. - Michel Marcus, Jul 04 2015

Triangle of coefficients in expansion of (4+x)^n. - N-E. Fahssi, Apr 13 2008

Set partitions of the n-set into at most 6 parts; also restricted growth strings (RGS) with six letters s(1),s(2),...,s(6) where the first occurrence of s(j) precedes the first occurrence of s(k) for all j < k. - Joerg Arndt, Jul 06 2011

Permuting the alphabet will not change a word structure. Thus aabc and bbca have the same structure.

Density of regular language L over {1,2,3,4,5,6}^* (i.e., number of strings of length n in L) described by regular expression with c=6: Sum_{i=1..c} Product_{j=1..i} (j(1+...+j)*) where Sum stands for union and Product for concatenation. - Nelma Moreira, Oct 10 2004

Word structures of length n using an N-ary alphabet are generated by taking M^n* the vector [(N 1's),0,0,0,...], leftmost column term = a(n+1). In the case of A056273, the vector = [1,1,1,1,1,1,0,0,0,...]. As the vector approaches all 1's, the leftmost column terms approach A000110, the Bell sequence. - Gary W. Adamson, Jun 23 2011

From Gary W. Adamson, Jul 06 2011: (Start)

Construct an infinite array of sequences representing word structures of length n using an N-ary alphabet as follows:

.

1, 1, 1, 1, 1, 1, 1, 1, ...; N=1, A000012

1, 2, 4, 8, 16, 32, 64, 128, ...; N=2, A000079

1, 2, 5, 14, 41, 122, 365, 1094, ...; N=3, A007051

1, 2, 5, 15, 51, 187, 715, 2795, ...; N=4, A007581

1, 2, 5, 15, 52, 202, 855, 3845, ...; N=5, A056272

1, 2, 5, 15, 52, 203, 876, 4111, ...; N=6, A056273

...

The sequences tend to A000110. Finite differences of columns reinterpreted as rows generate A008277 as a triangle: (1; 1,1; 1,3,1; 1,7,6,1; ...). (End)

Conjecture: a(n)/n tends to log(3)/(2*log(2)) = 0.792481250... (A094148). - Ed Pegg Jr, Dec 05 2002

Senge & Straus prove that for every m, there is some N such that for all n > N, a(n) > m. Dimitrov & Howe make this effective, proving that for n > 25, a(n) > 22. - Charles R Greathouse IV, Aug 23 2021

Ed Pegg's conjecture means that about half of the bits of 3^n are nonzero. It appears that the same is true for 5^n (A000351, cf. A118738) and 7^n (A000420). - M. F. Hasler, Apr 17 2024

Starting at 1, three-fold convolution of A000351 (powers of 5).

This is essentially A005117 (the squarefree numbers) raised to the fourth power. - T. D. Noe, Mar 13 2013

All positive integers have a unique factorization into powers of squarefree numbers with distinct exponents that are powers of two. So every positive number is a product of at most one squarefree number (A005117), at most one square of a squarefree number (A062503), at most one 4th power of a squarefree number (term of this sequence), at most one 8th power of a squarefree number, and so on. - Peter Munn, Mar 12 2020

Probably finite. Is 34694469519536141888238489627838134765625 the largest term?

This sequence is a permutation of A050376, so every positive integer is the product of a unique subset, S_factors, of its terms. If we restrict S_factors to be chosen from a subset, S_0, consisting of numbers from specified rows and/or columns of this array, there are notable sequences among those that may be generated. See the examples. Other notable sequences can be generated if we restrict the intersection of S_factors with specific rows/columns to have even cardinality. In any of the foregoing cases, the numbers in the resulting sequence form a group under the binary operation A059897(.,.).

Shares with array A246278 the property that columns grow downward by iterating A003961, and indeed, this array can be obtained from A246278 by selecting its columns 1, 2, 8, 128, ..., 2^((2^k)-1), for k >= 0.

A(n,k) is the image of the lattice point with coordinates X=n and Y=k under the inverse of the bijection f defined in the first comment of A306697. This geometric relationship can be used to construct an isomorphism from the polynomial ring GF(2)[x,y] to a ring over the positive integers, using methods similar to those for constructing A297845 and A306697. See A329329, the ring's multiplicative operator, for details.