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.

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A352321 Numbers k such that k and k+1 are both Pell-Niven numbers (A352320).

Original entry on oeis.org

1, 4, 5, 9, 14, 28, 29, 33, 39, 63, 87, 110, 111, 115, 125, 140, 164, 168, 169, 183, 255, 275, 308, 338, 410, 444, 483, 507, 564, 579, 584, 704, 791, 984, 985, 999, 1004, 1024, 1025, 1115, 1134, 1154, 1164, 1211, 1265, 1308, 1323, 1351, 1395, 1415, 1424, 1491
Offset: 1

Views

Author

Amiram Eldar, Mar 12 2022

Keywords

Comments

All the odd-indexed Pell numbers (A001653) are terms.

Examples

			4 is a term since 4 and 5 are both Pell-Niven numbers: the minimal Pell representation of 4, A317204(20) = 20, has the sum of digits 2+0 = 2 and 4 is divisible by 2, and the minimal Pell representation of 5, A317204(5) = 100, has the sum of digits 1+0+0 = 1 and 5 is divisible by 1.
		

Crossrefs

Programs

  • Mathematica
    pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; q[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; Divisible[n, Plus @@ IntegerDigits[ Total[3^(s - 1)], 3]]]; Select[Range[1500], q[#] && q[#+1] &]

A352343 Numbers k such that k and k+1 are both lazy-Pell-Niven numbers (A352342).

Original entry on oeis.org

1, 24, 63, 209, 216, 459, 560, 584, 656, 729, 999, 1110, 1269, 1728, 1859, 1989, 2100, 2196, 2197, 2255, 2650, 2651, 2820, 3443, 3497, 4080, 4563, 5291, 5784, 5785, 5837, 5928, 6252, 6383, 7344, 7657, 7812, 8150, 8203, 8459, 8670, 8749, 9251, 9295, 9372, 9464, 9840, 9884
Offset: 1

Views

Author

Amiram Eldar, Mar 12 2022

Keywords

Comments

Numbers k such that A352340(k) | k and A352340(k+1) | k+1.

Examples

			24 is a term since 24 and 25 are both lazy-Pell-Niven numbers: the maximal Pell representation of 24, A352339(24) = 1210, has the sum of digits A352340(24) = 1+2+1+0 = 4 and 24 is divisible by 4, and the maximal Pell representation of 25, A352339(25) = 1211, has the sum of digits A352340(25) = 1+2+1+1 = 5 and 25 is divisible by 5.
		

Crossrefs

Programs

  • Mathematica
    pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; pellp[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; IntegerDigits[Total[3^(s - 1)], 3]]; lazyPellNivenQ[n_] := Module[{v = pellp[n]}, nv = Length[v]; i = 1; While[i <= nv - 2, If[v[[i]] > 0 && v[[i + 1]] == 0 && v[[i + 2]] < 2, v[[i ;; i + 2]] += {-1, 2, 1}; If[i > 2, i -= 3]]; i++]; i = Position[v, _?(# > 0 &)]; Divisible[n, Plus @@ v[[i[[1, 1]] ;; -1]]]]; Select[Range[10^4], lazyPellNivenQ[#] && lazyPellNivenQ[#+1] &]

A352509 Numbers k such that k and k+1 are both Catalan-Niven numbers (A352508).

Original entry on oeis.org

1, 4, 5, 9, 32, 44, 55, 56, 134, 144, 145, 146, 155, 184, 234, 324, 329, 414, 426, 429, 434, 455, 511, 512, 603, 636, 930, 1004, 1014, 1160, 1183, 1215, 1287, 1308, 1448, 1472, 1505, 1562, 1595, 1808, 1854, 1967, 1985, 1995, 2051, 2075, 2096, 2135, 2165, 2255
Offset: 1

Views

Author

Amiram Eldar, Mar 19 2022

Keywords

Examples

			4 is a term since 4 and 5 are both Catalan-Niven numbers: the Catalan representation of 4, A014418(20) = 20, has the sum of digits 2+0 = 2 and 4 is divisible by 2, and the Catalan representation of 5, A014418(5) = 100, has the sum of digits 1+0+0 = 1 and 5 is divisible by 1.
		

Crossrefs

Programs

  • Mathematica
    c[n_] := c[n] = CatalanNumber[n]; q[n_] := Module[{s = {}, m = n, i}, While[m > 0, i = 1; While[c[i] <= m, i++]; i--; m -= c[i]; AppendTo[s, i]]; Divisible[n, Plus @@ IntegerDigits[Total[4^(s - 1)], 4]]]; Select[Range[2300], q[#] && q[#+1] &]

A364217 Numbers k such that k and k+1 are both Jacobsthal-Niven numbers (A364216).

Original entry on oeis.org

1, 2, 3, 8, 11, 14, 15, 27, 32, 42, 43, 44, 45, 51, 56, 75, 86, 87, 92, 95, 99, 104, 125, 128, 135, 144, 155, 171, 176, 182, 183, 195, 204, 264, 267, 275, 287, 305, 344, 363, 375, 387, 428, 444, 455, 474, 497, 512, 524, 535, 544, 545, 552, 555, 581, 605, 623, 639
Offset: 1

Views

Author

Amiram Eldar, Jul 14 2023

Keywords

Comments

A001045(2*n+1) = A007583(n) = (2^(2*n+1) + 1)/3 is a term for n >= 0, since its representation is 2*n 1's, so A364215(A001045(2*n+1)) = 1 divides A001045(2*n+1), and the representation of A001045(2*n+1) + 1 = (2^(2*n+1) + 4)/3 is max(2*n-1, 0) 0's between 2 1's, so A364215(A001045(2*n+1) + 1) = 2 which divides (2^(2*n+1) + 4)/3.

Crossrefs

Programs

  • Mathematica
    consecJacobsthalNiven[kmax_, len_] := Module[{m = 1, c = Table[False, {len}], s = {}}, Do[c = Join[Rest[c], {Divisible[k, DigitCount[m, 2, 1]]}]; While[m++; OddQ[IntegerExponent[m, 2]]]; If[And @@ c, AppendTo[s, k - len + 1]], {k, 1, kmax}]; s]; consecJacobsthalNiven[640, 2]
  • PARI
    lista(kmax, len) = {my(m = 1, c = vector(len)); for(k = 1, kmax, c = concat(vecextract(c, "^1"), !(k % sumdigits(m, 2))); until(valuation(m, 2)%2 == 0, m++); if(vecsum(c) == len, print1(k-len+1, ", ")));}
    lista(640, 2)

A364380 Numbers k such that k and k+1 are both greedy Jacobsthal-Niven numbers (A364379).

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 9, 10, 11, 14, 15, 20, 21, 26, 27, 32, 42, 43, 44, 45, 51, 56, 68, 75, 84, 85, 86, 87, 92, 99, 104, 105, 111, 115, 116, 125, 128, 135, 144, 155, 170, 171, 176, 182, 183, 195, 204, 213, 219, 224, 260, 264, 267, 275, 304, 305, 324, 329, 341, 344
Offset: 1

Views

Author

Amiram Eldar, Jul 21 2023

Keywords

Comments

The positive Jacobsthal numbers, A001045(n) for n >= 1, are terms since their representation in Jacobsthal greedy base is one 1 followed by n-1 0's, so A265745(A001045(n)) = 1 divides A001045(n), and the representation of A001045(n) + 1 is 2 if n <= 2 and otherwise n-3 0's between two 1's, so A265745(A001045(n) + 1) = 2 which divides A001045(n) + 1.

Crossrefs

Programs

  • Mathematica
    consecGreedyJN[kmax_, len_] := Module[{m = 1, c = Table[False, {len}], s = {}}, Do[c = Join[Rest[c], {greedyJacobNivenQ[k]}]; If[And @@ c, AppendTo[s, k - len + 1]], {k, 1, kmax}]; s]; consecGreedyJN[350, 2] (* using the function greedyJacobNivenQ[n] from A364379 *)
  • PARI
    lista(kmax, len) = {my(c = vector(len)); for(k = 1, kmax, c = concat(vecextract(c, "^1"), isA364379(k)); if(vecsum(c) == len, print1(k-len+1, ", ")));} \\ using the function isA364379(n) from A364379
    lista(350, 2)

A364007 Numbers k such that k and k+1 are both Wythoff-Niven numbers (A364006).

Original entry on oeis.org

3, 6, 7, 20, 39, 51, 54, 55, 90, 135, 143, 294, 305, 321, 356, 365, 369, 374, 375, 376, 784, 800, 924, 978, 979, 980, 986, 1904, 1945, 1970, 2043, 2199, 2232, 2289, 2394, 2424, 2439, 2499, 2525, 2562, 2580, 2583, 4185, 4598, 4707, 4774, 4790, 4796, 4879, 5004
Offset: 1

Views

Author

Amiram Eldar, Jul 01 2023

Keywords

Comments

A035508(n) = Fibonacci(2*n+2) - 1 is a term for n >= 2 since A135818(Fibonacci(2*n+2) - 1) = A135818(Fibonacci(2*n+2)) = 1.

Crossrefs

Programs

  • Mathematica
    seq[count_, nConsec_] := Module[{cn = wnQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {wnQ[k]}]; k++]; s]; seq[50, 2] (* using the function wnQ[n] from A364006 *)

A364124 Numbers k such that k and k+1 are both Stolarsky-Niven numbers (A364123).

Original entry on oeis.org

8, 56, 84, 159, 195, 224, 384, 399, 405, 995, 1140, 1224, 1245, 1295, 1309, 1419, 1420, 1455, 1474, 1507, 2585, 2597, 2600, 2680, 2681, 2727, 2744, 2750, 2799, 2855, 3122, 3311, 3339, 3345, 3618, 3707, 3795, 4004, 6770, 6774, 6984, 6985, 7014, 7074, 7154, 7405
Offset: 1

Views

Author

Amiram Eldar, Jul 07 2023

Keywords

Crossrefs

Programs

  • Mathematica
    seq[count_, nConsec_] := Module[{cn = stolNivQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {stolNivQ[k]}]; k++]; s]; seq[50, 2] (* using the function stolNivQ[n] from A364123 *)
  • PARI
    lista(count, nConsec) = {my(cn = vector(nConsec, i, isStolNivQ(i)), c = 0, k = nConsec + 1); while(c < count, if(vecsum(cn) == nConsec, c++; print1(k-nConsec, ", ")); cn = concat(vecextract(cn, "^1"), isStolNivQ(k)); k++);} \\ using the function isA364123(n) from A364123
    lista(50, 2)

A347495 Factorial base Niven numbers (A118363) with a record gap to the next factorial base Niven number.

Original entry on oeis.org

1, 2, 9, 12, 30, 40, 60, 192, 224, 318, 550, 640, 1136, 1989, 4875, 4980, 23355, 24272, 24378, 40131, 60192, 63872, 80472, 238680, 280140, 2027340, 2872620, 3622068, 13400475, 21293094, 25399080, 28584626, 111020840, 278690360, 355419734, 398884590, 834592590
Offset: 1

Views

Author

Amiram Eldar, Sep 03 2021

Keywords

Comments

The corresponding gaps are 1, 2, 3, 4, 5, 8, 10, 12, 16, 18, 20, 32, 34, 39, 52, 55, 60, 67, 82, 85, 90, 96, 154, 174, 210, 216, 222, 268, 297, 318, 336, 346, 430, 466, 517, 546, 604, ...

Examples

			The first 8 factorial base Niven numbers are 1, 2, 4, 6, 8, 9, 12 and 16. The gaps between them are 1, 2, 2, 2, 1, 3 and 4. The record gaps, 1, 2, 3 and 4, occur after the terms 1, 2, 9 and 12.
		

Crossrefs

Programs

  • Mathematica
    fivenQ[n_] := Module[{s = 0, i = 2, k = n}, While[k > 0, k = Floor[n/i!]; s = s + (i - 1)*k; i++]; Divisible[n, n - s]]; gapmax = 0; n1 = 1; s = {}; Do[If[fivenQ[n], gap = n - n1; If[gap > gapmax, gapmax = gap; AppendTo[s, n1]]; n1 = n], {n, 2, 10^5}]; s (* after Jean-François Alcover at A034968 *)

A377455 Numbers k such that k and k+1 are both terms in A377385.

Original entry on oeis.org

1, 1224, 126191, 428519, 649727, 1015416, 1988064, 3425856, 4542740, 4574240, 4743900, 4813668, 5131008, 6899840, 7001315, 7172424, 7356096, 8020583, 10206000, 11146421, 11566800, 11597999, 11693807, 12556700, 13742624, 13745759, 13831487, 14365120, 16939799, 20561400
Offset: 1

Views

Author

Amiram Eldar, Oct 29 2024

Keywords

Examples

			1224 is a term since both 1224 and 1225 are in A377385: 1224/A034968(1224) = 204 and 204/A034968(204) = 34 are integers, and 1225/A034968(1225) = 175 and 175/A034968(175) = 35 are integers.
		

Crossrefs

Cf. A034968.
Subsequence of A118363, A328205 and A377385.
Subsequences: A377456, A377457.
Analogous sequences: A376793 (binary), A377271 (Zeckendorf).

Programs

  • Mathematica
    fdigsum[n_] := Module[{k = n, m = 2, r, s = 0}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s += r; m++]; s]; q[k_] := q[k] = Module[{f = fdigsum[k]}, Divisible[k, f] && Divisible[k/f, fdigsum[k/f]]]; Select[Range[2*10^6], q[#] && q[#+1] &]
  • PARI
    fdigsum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, s += r; m++); s;}
    is1(k) = {my(f = fdigsum(k)); !(k % f) && !((k/f) % fdigsum(k/f));}
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}

A377457 Numbers k such that k and k+1 are both terms in A377386.

Original entry on oeis.org

1, 12563307224, 15897851550, 30412355999, 37706988600, 52576459775, 67673545631, 118533901904, 244316235000, 297265003100, 332110595000, 340800265728, 349358409503, 375624917760, 378624889440, 416375389115, 450026519903, 561162864248, 596004199840, 728643460544
Offset: 1

Views

Author

Amiram Eldar, Oct 29 2024

Keywords

Examples

			12563307224 is a term since both 12563307224 and 12563307225 are in A377386: 12563307224/A034968(12563307224) = 369509036, 369509036/A034968(369509036) = 9723922 and 9723922/A034968(9723922) = 373997 are integers, and 12563307225/A034968(12563307225) = 358951635, 358951635/A034968(358951635) = 7976703 and 7976703/A034968(7976703) = 257313 are integers.
		

Crossrefs

Cf. A034968.
Subsequence of A118363, A328205, A377385, A377386 and A377455.
Analogous sequences: A376795 (binary), A377272 (Zeckendorf).

Programs

  • PARI
    fdigsum(n) = {my(k = n, m = 2, r, s = 0); while([k, r] = divrem(k, m); k != 0 || r != 0, s += r; m++); s;}
    is1(k) = {my(f = fdigsum(k), f2, m); if(k % f, return(0)); m = k/f; f2 = fdigsum(m); !(m % f2) && !((m/f2) % fdigsum(m/f2));}
    lista(kmax) = {my(q1 = is1(1), q2); for(k = 2, kmax, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}
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