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.

Showing 1-4 of 4 results.

A340232 a(n) is the least number with exactly 2*n bi-unitary divisors.

Original entry on oeis.org

2, 6, 32, 24, 512, 96, 8192, 120, 131072, 1536, 2097152, 480, 33554432, 24576, 536870912, 840, 8589934592, 7776, 137438953472, 7680, 2199023255552, 6291456, 35184372088832, 3360, 562949953421312, 100663296, 9007199254740992, 122880, 144115188075855872, 124416
Offset: 1

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Author

Amiram Eldar, Jan 01 2021

Keywords

Comments

Every integer except 1 has an even number of bi-unitary divisors.

Examples

			a(1) = 2 since 2 is the least number with 2*1 = 2 bi-unitary divisors, 1 and 2.
a(2) = 6 since 6 is the least number with 2*2 = 4 bi-unitary divisors, 1, 2, 3 and 6.

Links

Crossrefs

Subsequence of A025487.
Cf. A222266, A286324, A293185.
Similar sequences: A005179 (all divisors), A038547 (odd divisors), A085629 (coreful divisors), A309181 (nonunitary), A340233 (exponential).

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e], e + 1, e]; d[1] = 1; d[n_] := Times @@ (f @@@ FactorInteger[n]);  max = 10; s = Table[0, {max}]; c = 0; n = 2;  While[c < max, i = d[n]/2; If[i <= max && s[[i]] == 0, c++; s[[i]] = n]; n++]; s

Formula

A286324(a(n)) = 2*n and A286324(k) != 2*n for all k < a(n).

A358252 a(n) is the least number with exactly n non-unitary square divisors.

Original entry on oeis.org

1, 8, 32, 128, 288, 864, 1152, 2592, 4608, 13824, 10368, 20736, 28800, 41472, 64800, 279936, 115200, 331776, 345600, 663552, 259200, 1679616, 518400, 1620000, 1166400, 4860000, 1036800, 17915904, 2073600, 15552000, 6998400, 26873856, 4147200, 53747712, 8294400
Offset: 0

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Author

Amiram Eldar, Nov 05 2022

Keywords

Comments

a(n) is the least number k such that A056626(k) = n.
Since A056626(k) depends only on the prime signature of k, all the terms of this sequence are in A025487.

Examples

			a(1) = 8 since 8 is the least number that has exactly one non-unitary square divisor, 4.

Links

Crossrefs

Cf. A056626, A025487, A358253.
Similar sequences: A005179 (all divisors), A038547 (odd divisors), A085629 (coreful divisors), A130279 (square), A187941 (even), A309181 (non-unitary), A340232 (bi-unitary), A340233 (exponential), A357450 (odd square).

Programs

  • Mathematica
    f1[p_, e_] := 1 + Floor[e/2]; f2[p_, e_] := 2^(1 - Mod[e, 2]); f[1] = 0; f[n_] := Times @@ f1 @@@ (fct = FactorInteger[n]) - Times @@ f2 @@@ fct; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[21, 10^6]
  • PARI
    s(n) = {my(f = factor(n)); prod(i = 1, #f~, 1 + floor(f[i,2]/2)) - 2^sum(i = 1, #f~, 1 - f[i,2]%2);}
lista(len, nmax) = {my(v = vector(len), c = 0, n = 1, i); while(c < len && n < nmax, i = s(n) + 1; if(i <= len && v[i] == 0, c++; v[i] = n); n++); v};

A358262 a(n) is the least number with exactly n noninfinitary square divisors.

Original entry on oeis.org

1, 16, 144, 256, 3600, 1296, 2304, 65536, 129600, 16777216, 32400, 20736, 57600, 331776, 589824, 4294967296, 6350400, 1099511627776, 150994944, 810000, 1587600, 1679616, 518400, 5308416, 2822400, 84934656, 8294400, 26873856, 14745600, 21743271936, 38654705664
Offset: 0

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Author

Amiram Eldar, Nov 06 2022

Keywords

Comments

a(n) is the least number k such that A358261(k) = n.
Since A358261(k) depends only on the prime signature of k, all the terms of this sequence are in A025487.

Examples

			a(1) = 16 since 16 is the least number with exactly one noninfinitary divisor, 4.

Links

Crossrefs

Cf. A025487, A358261, A358263.
Similar sequences: A005179 (all divisors), A038547 (odd divisors), A085629 (coreful divisors), A130279 (square), A187941 (even), A309181 (non-unitary), A340232 (bi-unitary), A340233 (exponential), A357450 (odd square), A358252 (non-unitary square).

Programs

  • Mathematica
    f1[p_, e_] := 1 + Floor[e/2]; f2[p_, e_] := 2^DigitCount[If[OddQ[e], e - 1, e], 2, 1]; f[1] = 0; f[n_] := Times @@ f1 @@@ (fct = FactorInteger[n]) - Times @@ f2 @@@ fct; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[15, 2*10^7]
  • PARI
    s(n) = {my(f = factor(n));  prod(i=1, #f~, 1+f[i,2]\2) - prod(i=1, #f~, 2^hammingweight(if(f[i,2]%2, f[i,2]-1, f[i,2])))};
lista(len, nmax) = {my(v = vector(len), c = 0, n = 1, i); while(c < len && n < nmax, i = s(n) + 1; if(i <= len && v[i] == 0, c++; v[i] = n); n++); v};

A361418 a(n) is the least number with exactly n noninfinitary divisors.

Original entry on oeis.org

1, 4, 12, 16, 60, 36, 48, 256, 360, 4096, 180, 144, 240, 576, 768, 65536, 2520, 1048576, 12288, 900, 1260, 1296, 720, 2304, 1680, 9216, 2880, 5184, 3840, 147456, 196608, 36864, 27720, 46656, 3145728, 4398046511104, 61440, 3600, 6300, 18014398509481984, 10080, 20736
Offset: 0

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Author

Amiram Eldar, Mar 11 2023

Keywords

Comments

a(n) is the least number k such that A348341(k) = n.
Since A348341(k) depends only on the prime signature of k, all the terms of this sequence are in A025487.

Examples

			a(1) = 4 since 4 is the least number with exactly one noninfinitary divisor, 2.

Links

Crossrefs

Cf. A025487, A348341, A348342.
Similar sequences: A005179 (all divisors), A038547 (odd divisors), A085629 (coreful divisors), A130279 (square), A187941 (even), A309181 (non-unitary), A340232 (bi-unitary), A340233 (exponential), A357450 (odd square), A358252 (non-unitary square).

Programs

  • Mathematica
    f[1] = 0; f[n_] := DivisorSigma[0, n] - Times @@ Flatten[2^DigitCount[#, 2, 1] & /@ FactorInteger[n][[;; , 2]]];
seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s];
seq[35, 10^7]
  • PARI
    s(n) = {my(f = factor(n)); numdiv(f) - prod(i = 1, #f~, 2^hammingweight(f[i,2]));}
lista(len, nmax) = {my(v = vector(len), c = 0, n = 1, i); while(c < len && n < nmax, i = s(n) + 1; if(i <= len && v[i] == 0, c++; v[i] = n); n++); v};
Showing 1-4 of 4 results.

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