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 1-9 of 9 results.

A109031 Numbers that have exactly eleven prime factors counted with multiplicity (A069272) whose digit reversal is different and also has 11 prime factors (with multiplicity).

Original entry on oeis.org

295245, 426816, 542592, 618624, 2112480, 2116224, 2150064, 2154816, 2196000, 2302560, 2327616, 2342277, 2388672, 2555280, 2576896, 2599200, 2768832, 2952288, 2952576, 4017216, 4074240, 4074840, 4076160, 4076568, 4078848
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 11 instance of the series which begins with k = 1 (emirps), k = 2, k = 3 (A109023), k = 4 (A109024), k = 5 (A109025), k = 6 (A109026), k = 7 (A109027), k = 8 (A109028), k = 9 (A109029), k = 10 (A109030).

Examples

			a(1) = 295245 is in this sequence because 295245 = 3^10 * 5 has exactly 11 prime factors counted with multiplicity and reverse(295245) = 542592 = 2^7 * 3^3 * 157 also has 11 prime factors counted with multiplicity.
		

Crossrefs

Programs

  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 11 && bigomega(r) == 11
    } \\ David A. Corneth, Mar 07 2024

Extensions

a(5)-a(25) from Donovan Johnson, Apr 09 2010

A109018 Least number with exactly n prime factors counted with multiplicity which gives a different number with exactly n prime factors counted with multiplicity when digits are reversed.

Original entry on oeis.org

13, 15, 117, 126, 270, 2576, 8820, 16560, 21168, 46848, 295245, 441600, 846720, 4078080, 80663040, 40590720, 2173236480, 4011724800, 21122906112, 40915058688, 274148425728, 63769149440, 2707602702336, 6167442456576, 21586195906560, 29798871072768, 420127895977984, 631722992467968
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

An emirp ("prime" spelled backwards) is a prime whose (base 10) reversal is also prime, but which is not a palindromic prime. The first few are 13, 17, 31, 37, 71, 73, 79, 97, 107, 113, 149, 157, ... (A006567).
An emirpimes ("semiprime" spelled backwards) is a semiprime whose (base 10) reversal is a different semiprime. A list of the first emirpimeses (or "semirpimes") are 15, 26, 39, 49, 51, 58, 62, 85, 93, 94, 115, 122, 123, ... (A097393).
An "emirp tsomla-3" ("3-almost prime" spelled backwards) is the k=3 sequence of the series for which k=1 are emirps and k=2 are emirpimes, a list of these being A109023. The union of these for k=1 through k = 13 is A109019.
The primes correspond to the "1-almost prime" numbers 2, 3, 5, 7, 11, ... (A000040). The 2-almost prime numbers correspond to semiprimes 4, 6, 9, 10, 14, 15, 21, 22, ... (A001358).
The first few 3-almost primes are 8, 12, 18, 20, 27, 28, 30, 42, 44, 45, 50, 52, 63, 66, 68, 70, 75, 76, 78, 92, 98, 99, ... (A014612). The first few 4-almost primes are 16, 24, 36, 40, 54, 56, 60, 81, 84, 88, 90, 100, ... (A014613).
The first few 5-almost primes are 32, 48, 72, 80, ... (A014614).
The Mathematica code for this was written by Ray Chandler, who has coauthorship credit for this sequence.

Examples

			a(1) = 13 because 13 is the smallest "emirp" (prime which, digits reversed, becomes a different prime) since reverse(13) = 31 is prime.
a(2) = 15 because 15 is the smallest emirpimes ("semiprime" spelled backwards) as a semiprime whose (base 10) reversal is a different semiprime. The first such number is 15, since 15 reversed is 51 and both 15 and 51 are semiprimes (i.e. 15 = 3 * 5 and 51 = 3 * 17).
a(3) = 117 because 117 is the smallest "emirp tsomla-3" ("3-almost prime" spelled backwards) since 117 reversed is 711 and 117 = 3^2 * 13 and 711 = 3^2 * 79.
		

Crossrefs

Programs

  • Mathematica
    kAlmost[n_] := Plus @@ Last /@ FactorInteger@n; fQ[n_] := Block[{id = IntegerDigits@n, k = kAlmost@n}, If[id != Reverse@id && k == kAlmost@FromDigits@Reverse@id, k, -1]]; t = Table[0, {20}]; Do[ a = fQ@n; If[a < 20 && t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 10, 150000000}] (* Robert G. Wilson v, Jan 06 2008 *)
    Table[Select[Range[41*10^5],!PalindromeQ[#]&&PrimeOmega[#]==PrimeOmega[ IntegerReverse[ #]] ==n&][[1]],{n,14}] (* The program generates the first 14 terms of the sequence. *) (* Harvey P. Dale, Oct 15 2023 *)

Extensions

a(14)-a(16) from Robert G. Wilson v, Jan 06 2008
a(17)-a(24) from Donovan Johnson, Nov 17 2008
a(25)-a(28) from Michael S. Branicky, Jun 04 2024

A109023 3-almost primes (A014612) whose digit reversal is different and also has 3 prime factors (with multiplicity).

Original entry on oeis.org

117, 147, 165, 244, 246, 285, 286, 290, 338, 366, 369, 406, 418, 425, 435, 438, 442, 475, 498, 506, 507, 508, 524, 534, 539, 548, 561, 574, 582, 604, 605, 609, 628, 642, 663, 670, 682, 705, 711, 741, 759, 805, 814, 826, 833, 834, 845, 890, 894, 906, 935
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 3 instance of the series which begins with k = 1, k = 2.

Examples

			1066 is in this sequence because 1066 = 2 * 13 * 41, making it a 3-almost prime and reverse(1066) = 6601 = 7 * 23 * 41, also a 3-almost prime.
2001 is in this sequence because 2001 = 3 * 23 * 29 and reverse(2001) = 1002 = 2 * 3 * 167.
		

References

  • W. W. R. Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th ed. New York: Dover, pp. 14-15, 1987.
  • J. Edalj, Problem 1622. L'Intermédiaire des Mathématiciens, 16, 34, 1909.

Crossrefs

Programs

  • Mathematica
    Select[Range[1000],PrimeOmega[#]==3&&PrimeOmega[FromDigits[Reverse[IntegerDigits[#]]]]==3&&!PalindromeQ[#]&] (* James C. McMahon, Mar 06 2024 *)
  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 3 && bigomega(r) == 3
    } \\ David A. Corneth, Mar 07 2024

Extensions

1002 replaced by 935 - R. J. Mathar, Dec 14 2009

A109024 Numbers that have exactly four prime factors counted with multiplicity (A014613) whose digit reversal is different and also has 4 prime factors (with multiplicity).

Original entry on oeis.org

126, 225, 294, 315, 459, 488, 492, 513, 522, 558, 621, 650, 738, 837, 855, 884, 954, 1035, 1062, 1098, 1107, 1197, 1206, 1236, 1287, 1305, 1422, 1518, 1617, 1665, 1917, 1926, 1956, 1962, 1989, 2004, 2034, 2046, 2068, 2104, 2148, 2170, 2180, 2223, 2226
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 4 instance of the series which begins with k = 1, k = 2, k = 3 (A109023).

Examples

			a(1) = 126 is in this sequence because 126 = 2 * 3^2 * 7 is a 4-almost prime and reverse(126) = 621 = 3^3 * 23 is also a 4-almost prime.
a(2) = 225 is in this sequence because 225 = 3^2 * 5^2 is a 4-almost prime and reverse(225) = 522 = 2 * 3^2 * 29 is also a 4-almost prime. (That 225 and 522 are concatenated from entirely prime digits is a coincidence, as with 2223).
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2226],PrimeOmega[#]==4 && PrimeOmega[FromDigits[Reverse[IntegerDigits[#]]]]==4 &&!PalindromeQ[#]&] (* James C. McMahon, Mar 07 2024 *)
  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 4 && bigomega(r) == 4
    } \\ David A. Corneth, Mar 07 2024

A109025 Numbers that have exactly five prime factors counted with multiplicity (A014614) whose digit reversal is different and also has 5 prime factors (with multiplicity).

Original entry on oeis.org

270, 1386, 1575, 2070, 2136, 2142, 2295, 2300, 2394, 2412, 2475, 2508, 2550, 2565, 2568, 2610, 2844, 2964, 3087, 3267, 3465, 3654, 3708, 3924, 4008, 4016, 4068, 4185, 4208, 4290, 4293, 4347, 4446, 4482, 4563, 4692, 4779, 4875, 4932, 5049, 5238, 5355
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 5 instance of the series which begins with k = 1, k = 2, k = 3 (A109023), k = 4 (A109024).

Examples

			a(2) = 1386 is in this sequence because 1386 = 2 * 3^2 * 7 * 11 has exactly 5 prime factors counted with multiplicity and reverse(1386) = 6831 = 3^3 * 11 * 23 is also has exactly 5 prime factors counted with multiplicity.
5355 is in this sequence because 5355 = 3^2 * 5 * 7 * 17 and reverse(5355) = 5535 = 3^3 * 5 * 41.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[6000],!PalindromeQ[#]&&Total[FactorInteger[#][[All,2]]]==Total[ FactorInteger[ IntegerReverse[#]][[All,2]]]==5&] (* Harvey P. Dale, Nov 20 2022 *)
  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 5 && bigomega(r) == 5
    } \\ David A. Corneth, Mar 07 2024

Extensions

Typo in definition corrected by Harvey P. Dale, Nov 20 2022

A109027 Numbers that have exactly seven prime factors counted with multiplicity (A046308) whose digit reversal is different and also has 7 prime factors (with multiplicity).

Original entry on oeis.org

8820, 21240, 21708, 21780, 21920, 23280, 23472, 23625, 23800, 25560, 25584, 25758, 26280, 27432, 27504, 27888, 27900, 28836, 29250, 29403, 29736, 29970, 30492, 34884, 36828, 40338, 40572, 40950, 41976, 42228, 42984, 43659, 43956, 44128
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 7 instance of the series which begins with k = 1 (emirps), k = 2, k = 3, k = 4, k = 5 (A109025), k = 6 (A109026).

Examples

			a(20) = 29403 is in this sequence because 29403 = 3^5 * 11^2 has exactly 7 prime factors counted with multiplicity and reverse(29403) = 30492 = 2^2 * 3^2 * 7 * 11^2 also has exactly 7 prime factors counted with multiplicity.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[45000],!PalindromeQ[#]&&PrimeOmega[#]==PrimeOmega[ IntegerReverse[ #]] ==7&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 02 2019 *)
  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 7 && bigomega(r) == 7
    } \\ David A. Corneth, Mar 07 2024

A109028 Numbers that have exactly eight prime factors counted with multiplicity (A046310) whose digit reversal is different and also has 8 prime factors (with multiplicity).

Original entry on oeis.org

16560, 25515, 27864, 42480, 46872, 51552, 57348, 61488, 65448, 67797, 69408, 69840, 79776, 80496, 84375, 84456, 88416, 105336, 119448, 125928, 160416, 167076, 202032, 204984, 206928, 210960, 211104, 211464, 213750, 213792, 213920, 213984
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 8 instance of the series which begins with k = 1 (emirps), k = 2, k = 3 (A109023), k = 4 (A109024), k = 5 (A109025), k = 6 (A109026), k = 7 (A109027).

Examples

			a(2) = 25515 is in this sequence because 25515 = 3^6 * 5 * 7 has exactly 8 prime factors counted with multiplicity and reverse(25515) = 51552 = 2^5 * 3^2 * 179 also has exactly 8 prime factors counted with multiplicity.
		

Crossrefs

Programs

  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 8 && bigomega(r) == 8
    } \\ David A. Corneth, Mar 07 2024

A109029 Numbers that have exactly nine prime factors counted with multiplicity (A046312) whose digit reversal is different and also has 9 prime factors (with multiplicity).

Original entry on oeis.org

21168, 23424, 23616, 27456, 41184, 42432, 48114, 61632, 65472, 86112, 211410, 212256, 213192, 215232, 217440, 219072, 230208, 232512, 236925, 236928, 238656, 238680, 251100, 251505, 251748, 253824, 255024, 255960, 257856, 259968, 270912
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 8 instance of the series which begins with k = 1 (emirps), k = 2, k = 3 (A109023), k = 4 (A109024), k = 5 (A109025), k = 6 (A109026), k = 7 (A109027), k = 8 (A109028).

Examples

			a(1) = 21168 is in this sequence because 21168 = 2^4 * 3^3 * 7^2 has exactly 9 prime factors counted with multiplicity and reverse(21168) = 86112 = 2^5 * 3^2 * 13 * 23 also has exactly 9 prime factors counted with multiplicity.
		

Crossrefs

Programs

  • Mathematica
    okQ[n_]:=Module[{idn=IntegerDigits[n],ridn},ridn=Reverse[idn];idn!=ridn && PrimeOmega[n]==9&&PrimeOmega[FromDigits[ridn]]==9]; Select[Range[ 271000],okQ] (* Harvey P. Dale, Sep 24 2011 *)
  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 9 && bigomega(r) == 9
    } \\ David A. Corneth, Mar 07 2024

A109030 Numbers that have exactly ten prime factors counted with multiplicity (A046314) whose digit reversal is different and also has 10 prime factors (with multiplicity).

Original entry on oeis.org

46848, 84864, 217152, 219456, 232848, 251712, 257664, 259776, 274104, 276048, 401472, 415584, 422820, 428160, 428736, 447360, 466752, 485514, 637824, 650160, 654912, 677952, 808320, 840672, 846369, 848232, 963648
Offset: 1

Views

Author

Jonathan Vos Post, Jun 16 2005

Keywords

Comments

This sequence is the k = 10 instance of the series which begins with k = 1 (emirps), k = 2, k = 3 (A109023), k = 4 (A109024), k = 5 (A109025), k = 6 (A109026), k = 7 (A109027), k = 8 (A109028), k = 9 (A109029).

Examples

			a(1) = 46848 is in this sequence because 46848 = 2^8 * 3 * 61 has exactly 10 prime factors counted with multiplicity and reverse(46848) = 84864 = 2^7 * 3 * 13 * 17 also has exactly 10 prime factors counted with multiplicity.
		

Crossrefs

Programs

  • Mathematica
    taQ[n_]:=Module[{idn=IntegerDigits[n],rev},rev=Reverse[idn];rev!=idn&&PrimeOmega[n] == 10 == PrimeOmega[FromDigits[rev]]]; Select[Range[ 1000000], taQ] (* Harvey P. Dale, May 03 2013 *)
  • PARI
    is(n) = {
    	my(r = fromdigits(Vecrev(digits(n))));
    	n!=r && bigomega(n) == 10 && bigomega(r) == 10
    } \\ David A. Corneth, Mar 07 2024
Showing 1-9 of 9 results.