A384211 -id:A384211 - cp's OEIS frontend

A384212 a(n) is the number of bases >= 2 in which the alternating sum of digits of n is equal to 0.

0, 0, 1, 1, 1, 2, 1, 2, 2, 2, 1, 4, 1, 2, 3, 3, 1, 4, 1, 3, 2, 2, 1, 6, 2, 2, 3, 4, 1, 6, 1, 4, 3, 2, 2, 7, 1, 2, 3, 6, 1, 5, 1, 4, 5, 2, 1, 8, 2, 3, 3, 4, 1, 5, 2, 6, 3, 2, 1, 9, 1, 2, 5, 5, 3, 6, 1, 4, 2, 6, 1, 10, 1, 2, 5, 4, 2, 5, 1, 8, 3, 2, 1, 8, 3, 2, 2

Offset: 1

Felix Huber, May 24 2025

The alternating sum of digits of n is equal to 1 for base n and equal to n for bases > n.

			a(72) = 10 because the alternating sum of digits of n is equal to 0 in the 10 bases 2 [1, 0, 0, 1, 0, 0, 0], 3 [2, 2, 0, 0], 5 [2, 4, 2], 7 [1, 3, 2], 8 [1, 1, 0], 11 [6, 6], 17 [4, 4], 23 [3, 3], 35 [2, 2] and 71 [1, 1].

Felix Huber, Table of n, a(n) for n = 1..10000

Cf. A055240, A061845, A225693, A135499, A135551, A384211.

A384212:=proc(n)
    local a,b,c,i;
    a:=0;
    for b from 2 to n-1 do
        c:=convert(n,'base',b);
    	   if add(c[i]*(-1)^i,i=1..nops(c))=0 then
           a:=a+1
        fi
    od;
    return a
end proc;
seq(A384212(n),n=1..87);
A384212bases:=proc(n)
    local L,b,c,i;
    L:=[];
    for b from 2 to n-1 do
        c:=convert(n,'base',b);
    	if add(c[i]*(-1)^i,i=1..nops(c))=0 then
            L:=[op(L),b,ListTools:-Reverse(c)]
        fi
    od;
    return op(L)
end proc;
A384212bases(72);

q[n_, b_] := Module[{d = IntegerDigits[n, b]}, Sum[(-1)^k*d[[k]], {k, 1, Length[d]}] == 0 ]; a[n_] := Count[Range[2, n-1], ?(q[n, #] &)]; Array[a, 100] (* _Amiram Eldar, May 24 2025 *)

a(n) = sum(b=2, n-1, my(d=digits(n, b)); sum(k=1, #d, (-1)^k*d[k]) == 0); \\ Michel Marcus, May 24 2025

from sympy.ntheory import digits
def s(v): return sum(v[::2]) - sum(v[1::2])
def a(n): return sum(1 for b in range(2, n) if s(digits(n, b)[1:]) == 0)
print([a(n) for n in range(1, 87)]) # Michael S. Branicky, May 24 2025

Trivial bounds: 1 <= a(n) <= n - 2 for n >= 3 because the representation of n in base n-1 is [1,1] and the alternating sum of digits of n is > 0 for bases >= n.

A384436 a(n) is the number of distinct ways to represent n in any integer base >= 2 using only square digits.

Original entry on oeis.org

1, 1, 1, 2, 4, 3, 3, 3, 3, 6, 5, 4, 5, 5, 4, 4, 6, 5, 4, 5, 7, 7, 5, 5, 7, 8, 6, 6, 8, 7, 7, 7, 7, 7, 7, 6, 11, 9, 6, 7, 10, 7, 7, 7, 8, 8, 8, 6, 8, 11, 7, 7, 9, 10, 7, 7, 10, 10, 7, 7, 11, 10, 7, 7, 13, 11, 7, 7, 11, 10, 7, 7, 10, 11, 8, 8, 11, 11, 9, 8, 11, 15

Offset: 0

Felix Huber, May 29 2025

The representations of n remain the same for bases greater than n, as they all consist solely of the digit n.

			The a(36) = 11 distinct ways to represent 36 using only square digits are [1,0,0,1,0,0] in base 2, [1,1,0,0] in base 3, [1,0,0] in base 6, [4,4] in base 8, [4,0] in base 9, [1,16] in base 20, [1,9] in base 27, [1,4] in base 32, [1,1] in base 35, [1,0] in base 36 and [36] in bases >= 37.

Felix Huber, Table of n, a(n) for n = 0..10000

Cf. A046030, A055240, A061845, A077268, A126071, A135551, A384211, A384212.

A384436:=proc(n)
    local a,b,c;
    a:=0;
    for b from 2 to n+1 do
        c:=convert(n,'base',b);
        if select(issqr,c)=c then
            a:=a+1
        fi
    od;
    return max(1,a)
end proc;
seq(A384436(n),n=0..81);

a[n_] := Sum[Boole[AllTrue[IntegerDigits[n, b], IntegerQ[Sqrt[#]] &]], {b, 2, n+1}]; a[0] = 1; Array[a, 100, 0] (* Amiram Eldar, May 29 2025 *)

a(n) = sum(b=2, n+1, my(d=digits(n,b)); #select(issquare, d) == #d); \\ Michel Marcus, May 29 2025

Trivial lower bound for n >= 2: a(n) >= 2 for nonsquares n and a(n) >= 3 for squares n because in base 2 the representations of n consists only of the square digits '0' and '1', in base n the representation of n is [1,0] and in bases > n the representation of n is [n].

cp's OEIS Frontend

A384212 a(n) is the number of bases >= 2 in which the alternating sum of digits of n is equal to 0.

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