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 1-8 of 8 results.

A352320 Pell-Niven numbers: numbers that are divisible by the sum of the digits in their minimal (or greedy) representation in terms of the Pell numbers (A317204).

Original entry on oeis.org

1, 2, 4, 5, 6, 9, 10, 12, 14, 15, 18, 20, 24, 28, 29, 30, 33, 34, 36, 39, 40, 42, 44, 48, 50, 58, 60, 63, 64, 68, 70, 72, 82, 84, 87, 88, 90, 92, 96, 110, 111, 112, 115, 116, 120, 125, 126, 135, 140, 141, 144, 155, 164, 165, 168, 169, 170, 174, 180, 183, 184, 186
Offset: 1

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Author

Amiram Eldar, Mar 12 2022

Keywords

Comments

Numbers k such that A265744(k) | k.
All the positive Pell numbers (A000129) are terms.

Examples

			6 is a term since its minimal Pell representation, A317204(6) = 101, has A265744(6) = 2 1's and 6 is divisible by 2.
		

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[200], q]

A352322 Starts of runs of 3 consecutive Pell-Niven numbers (A352320).

Original entry on oeis.org

4, 28, 110, 168, 984, 1024, 3123, 3514, 5740, 6783, 6923, 8584, 12664, 16744, 18160, 19670, 23190, 23470, 24030, 34503, 34643, 36304, 40384, 45880, 47390, 50910, 51190, 51750, 57607, 61640, 68104, 73600, 78403, 78630, 78910, 79470, 86674, 89360, 95824, 101320
Offset: 1

Views

Author

Amiram Eldar, Mar 12 2022

Keywords

Comments

Conjecture: There are no runs of 4 consecutive Pell-Niven numbers (checked up to 2*10^8).

Examples

			4 is a term since 4, 5 and 6 are all 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, the minimal Pell representation of 5, A317204(5) = 100, has the sum of digits 1+0+0 = 1 and 5 is divisible by 1, and the minimal Pell representation of 6, A317204(6) = 101, has the sum of digits 1+0+1 = 2 and 6 is divisible by 2.
		

Crossrefs

A182190 \ {0} is a subsequence.
Subsequence of A352320 and A352321.

Programs

  • Mathematica
    pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; pellNivenQ[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]]]; seq[count_, nConsec_] := Module[{pn = pellNivenQ /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ pn, c++; AppendTo[s, k - nConsec]]; pn = Join[Rest[pn], {pellNivenQ[k]}]; k++]; s]; seq[30, 3]

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)
Showing 1-8 of 8 results.