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.

Previous Showing 11-15 of 15 results.

A075467 Trajectory of 270798 under the Reverse and Add! operation carried out in base 4, written in base 10.

Original entry on oeis.org

270798, 1005135, 1994670, 5058075, 20047500, 33313725, 66545850, 112201785, 225464610, 368353785, 835135950, 1587633825, 2841028950, 5347819200, 5598498225, 10862757750, 21453946560, 22456662705, 43576370550
Offset: 0

Views

Author

Klaus Brockhaus, Sep 18 2002

Keywords

Comments

The base 4 trajectory of 270798 = A075421(370) provably does not contain a palindrome. A proof along the lines of Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2, can be based on the formula given below. - The generating function given describes the sequence from a(11) onward; the g.f. for the complete sequence is known but nearly twice as big.

Examples

			270798 (decimal) = 1002013032 -> 1002013032 + 2303102001 = 3311121033 = 1005135 (decimal).
		

Crossrefs

Programs

  • Mathematica
    NestWhileList[# + IntegerReverse[#, 4] &, 270798,  # !=
    IntegerReverse[#, 4] &, 1, 23] (* Robert Price, Oct 18 2019 *)
  • PARI
    {m=270798; stop=20; c=0; while(c0,d=divrem(k,4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

Formula

a(0), ..., a(10) as above; for n > 10 and n = 5 (mod 6): a(n) = 5*4^(2*k+10)+15341035*4^k-15 where k = (n+1)/6; n = 0 (mod 6): a(n) = 10*4^(2*k+10)+9792150*4^k-10 where k = n/6; n = 1 (mod 6): a(n) = 20*4^(2*k+10)-1305620*4^k where k = (n-1)/6; n = 2 (mod 6): a(n) = 20*4^(2*k+10)+14361820*4^k-15 where k = (n-2)/6; n = 3 (mod 6): a(n) = 40*4^(2*k+10)+7833720*4^k-10 where k = (n-3)/6; n = 4 (mod 6): a(n) = 80*4^(2*k+10)-1305620*4^k where k = (n-4)/6. G.f.: -15*(1426085120*x^11+749251744*x^10+419191024*x^9-1430263104*x^8-715827880*x^7-369055228*x^6-352343296*x^5-222825800*x^4-155978060*x^3+356521280*x^2+189401930*x+105842255)/((x-1)*(x^2+x+1)*(2*x^3-1)*(2*x^3+1)*(4*x^3-1))

A075299 Trajectory of 290 under the Reverse and Add! operation carried out in base 4, written in base 10.

Original entry on oeis.org

290, 835, 1610, 4195, 17060, 23845, 46490, 89080, 138125, 255775, 506510, 1238395, 5127260, 8616205, 15984335, 31949470, 79793675, 315404860, 569392925, 1060061935, 2114961710, 5206421995, 20997654620, 35262166285
Offset: 0

Views

Author

Klaus Brockhaus, Sep 12 2002

Keywords

Comments

290 is conjectured (cf. A066450) to be the smallest number such that the Reverse and Add! algorithm in base 4 does not lead to a palindrome. Unlike 318 (cf. A075153) its trajectory does not exhibit any recognizable regularity, so that the method by which the base 4 trajectory of 318 as well as the base 2 trajectories of 22 (cf. A061561), 77 (cf. A075253), 442 (cf. A075268) etc. can be proved to be palindrome-free (cf. Links), is not applicable here.

Examples

			290 (decimal) = 10202 -> 10202 + 20201 = 31003 = 835 (decimal).
		

Crossrefs

Programs

  • Mathematica
    NestWhileList[# + IntegerReverse[#, 4] &, 290,  # !=
    IntegerReverse[#, 4] &, 1, 23] (* Robert Price, Oct 18 2019 *)
  • PARI
    {m=290; stop=26; c=0; while(c0,d=divrem(k,4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

A076247 Trajectory of 1059774 under the Reverse and Add! operation carried out in base 4, written in base 10.

Original entry on oeis.org

1059774, 4187583, 8355006, 20822715, 83391660, 144328605, 268919295, 1339676160, 1349598705, 2683144950, 5361370860, 9358549725, 17380163775, 85563883200, 89574690225, 173801637750, 343262166720, 359352580785
Offset: 0

Views

Author

Klaus Brockhaus, Oct 03 2002

Keywords

Comments

1059774 = A075421(1096) is the fourth term of A075421 whose base 4 trajectory provably does not contain a palindrome. A proof along the lines of Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2, can be based on the formula given below. - The generating function given describes the sequence from a(16) onward; the g.f. for the complete sequence is known but nearly twice as big.

Examples

			1059774 (decimal) = 10002232332 -> 10002232332 + 23323220001 = 33332112333 = 4187583 (decimal).
		

Crossrefs

Programs

  • Mathematica
    NestWhileList[# + IntegerReverse[#, 4] &, 1059774,  # !=
    IntegerReverse[#, 4] &, 1, 23] (* Robert Price, Oct 18 2019 *)
  • PARI
    {m=1059774; stop=19; c=0; while(c0,d=divrem(k,4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

Formula

a(0), ..., a(15) as above; for n > 15 and n = 4 (mod 6): a(n) = 5*4^(2*k+12)-5237765*4^k where k = (n+2)/6; n = 5 (mod 6): a(n) = 5*4^(2*k+12)+246174955*4^k-15 where k = (n+1)/6; n = 0 (mod 6): a(n) = 10*4^(2*k+12)+157132950*4^k-10 where k = n/6; n = 1 (mod 6): a(n) = 20*4^(2*k+12)-20951060*4^k where k = (n-1)/6; n = 2 (mod 6): a(n) = 20*4^(2*k+12)+230461660*4^k-15 where k = (n-2)/6; n = 3 (mod 6): a(n) = 40*4^(2*k+12)+125706360*4^k-10 where k = (n-3)/6. G.f.: -15*(185397326496*x^11+95559181296*x^10+91268404224*x^9-183251937960*x^8-92341098492*x^7 -91268404224*x^6-48628806952*x^5-27174921532*x^4-22884144448*x^3+46483418410*x^2 +23956838719*x+22884144448)/((x-1)*(x^2+x+1)*(2*x^3-1)*(2*x^3+1)*(4*x^3-1))

A076248 Trajectory of 1059831 under the Reverse and Add! operation carried out in base 4, written in base 10.

Original entry on oeis.org

1059831, 4728312, 7831065, 14433270, 24913965, 56412450, 92165625, 208908750, 396926625, 710289750, 1336954560, 1398889905, 2715199350, 5363547840, 5614238385, 10894222710, 21453945600, 21701687025, 43073052150
Offset: 0

Views

Author

Klaus Brockhaus, Oct 03 2002

Keywords

Comments

1059831 = A075421(1105 ) is the fifth term of A075421 whose base 4 trajectory provably does not contain a palindrome. A proof along the lines of Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2, can be based on the formula given below.

Examples

			1059831 (decimal) = 10002233313 -> 10002233313 + 31333220001 = 102002113320 = 4728312 (decimal).
		

Crossrefs

Programs

  • Mathematica
    NestWhileList[# + IntegerReverse[#, 4] &, 1059831,  # != IntegerReverse[ #, 4] &, 1, 23] (* Robert Price, Oct 19 2019 *)
  • PARI
    {m=1059831; stop=19; c=0; while(c0,d=divrem(k,4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

Formula

a(0), ..., a(7) as above; for n > 7 and n = 2 (mod 6): a(n) = 5*4^(2*k+9)+3836395*4^k-15 where k = (n+4)/6; n = 3 (mod 6): a(n) = 10*4^(2*k+9)+2450070*4^k-10 where k = (n+3)/6; n = 4 (mod 6): a(n) = 20*4^(2*k+9)-326420*4^k where k = (n+2)/6; n = 5 (mod 6): a(n) = 20*4^(2*k+9)+3544540*4^k-15 where k = (n+1)/6; n = 0 (mod 6): a(n) = 40*4^(2*k+9)+1927800*4^k-10 where k = n/6; n = 1 (mod 6): a(n) = 80*4^(2*k+9)-322580*4^k where k = (n-1)/6. G.f.: -3*(668508000*x^19+444361200*x^18+222142800*x^17-528080680*x^16-356464620*x^15 -125753060*x^14-299532884*x^13-188180432*x^12-143040640*x^11+128992350*x^10+90219415*x^9 +38288125*x^8+28112975*x^7+6666425*x^6+5752375*x^5+424135*x^4+3044705*x^3+2610355*x^2 + 1576104*x+353277)/((x-1)*(x^2+x+1)*(2*x^3-1)*(2*x^3+1)*(4*x^3-1))

A077408 Trajectory of 103 under the Reverse and Add! operation carried out in base 3, written in base 10.

Original entry on oeis.org

103, 230, 436, 776, 2424, 3856, 7400, 20856, 30928, 60920, 220248, 242704, 432896, 857152, 1460408, 2754688, 5134016, 16206744, 24437488, 44623424, 138104472, 201737128, 401511824, 1438324704, 1601682040, 2820726320, 5622321088
Offset: 0

Views

Author

Klaus Brockhaus, Nov 05 2002

Keywords

Comments

103 = A077405(0) is conjectured (cf. A066450) to be the smallest number such that the Reverse and Add! algorithm in base 3 does not lead to a palindrome. Its trajectory does not exhibit any recognizable regularity, so that the method by which the base-2 trajectories of 22 (cf. A061561), 77 (cf. A075253), 442 (cf. A075268) etc. as well as the base-4 trajectories of 318 (cf. A075153), 266718 (cf. A075466), 270798 (cf. A075467) etc. can be proved to be palindrome-free (cf. Links), is not applicable here.

Examples

			103 (decimal) = 10211 -> 10211 + 11201 = 22112 = 230 (decimal).
		

Crossrefs

Programs

  • ARIBAS
    m := 103; stop := 28; c := 0; while c < stop do write(m:group(0),","); k := m; rev := 0; while k > 0 do rev := 3*rev + (k mod 3); k := k div 3; end; inc(c); m := m+rev; end;
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