A064671 -id:A064671 - cp's OEIS frontend

A064544 Biquanimous numbers (or biquams): group the digits into two pieces (not necessarily equal or in order) with the same sum.

0, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 110, 112, 121, 123, 132, 134, 143, 145, 154, 156, 165, 167, 176, 178, 187, 189, 198, 202, 211, 213, 220, 224, 231, 235, 242, 246, 253, 257, 264, 268, 275, 279, 286, 297, 303, 312, 314, 321, 325, 330, 336, 341, 347, 352, 358

Offset: 0

David W. Wilson, Oct 09 2001

This sequence is 10-automatic (decimal expansions form a regular language accepted by a finite automaton).

			143 is in the sequence because its digits {1, 4, 3} may be grouped so that 1+3 = 4.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Index entries for 10-automatic sequences.

Cf. A064671, A064686 (number of n-digit base-3 biquams), A065023, A065024, A065025.

is(n) = { my (d=digits(n), s=[0]); for (k=1, #d, s=setunion(apply(v -> v+d[k], s), apply(v -> v-d[k], s))); setsearch(s, 0)>0 } \\ Rémy Sigrist, Jan 23 2021

A288687 Number of n-digit biquanimous strings using digits {0,1,2,3}.

Original entry on oeis.org

1, 1, 4, 19, 92, 421, 1830, 7687, 31624, 128521, 518666, 2084875, 8361996, 33497101, 134094862, 536608783, 2146926608, 8588754961, 34357248018, 137433710611, 549744803860, 2199000186901, 8796044787734, 35184271425559, 140737278640152, 562949517213721

Offset: 0

Alois P. Heinz, Jun 13 2017

A biquanimous string is a string whose digits can be split into two groups with equal sums.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-37,64,-52,16).

Column k=3 of A288638.

LinearRecurrence[{10,-37,64,-52,16},{1,1,4,19,92,421},30] (* Harvey P. Dale, Jul 29 2017 *)

Vec((1 - 9*x + 31*x^2 - 48*x^3 + 38*x^4 - 16*x^5) / ((1 - x)^2*(1 - 2*x)^2*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Dec 16 2017

G.f.: (1 - 9*x + 31*x^2 - 48*x^3 + 38*x^4 - 16*x^5) / ((1 - x)^2*(1 - 2*x)^2*(1 - 4*x)).
a(n) = 1 + A064671(n) for n > 0.
From Colin Barker, Dec 16 2017: (Start)
a(n) = (2^(2*n-1) + n - 2^(n-1)*(1+n)).
a(n) = 10*a(n-1) - 37*a(n-2) + 64*a(n-3) - 52*a(n-4) + 16*a(n-5) for n>5.
(End)

cp's OEIS Frontend

A064544 Biquanimous numbers (or biquams): group the digits into two pieces (not necessarily equal or in order) with the same sum.

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