åtte - cp's OEIS frontend

User: åtte

åtte has authored 27 sequences. Here are the ten most recent ones:

A178055 Numbers representing the number of days in a month in the Gregorian calendar (modulus 7).

3, 1, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 1, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3, 3, 1, 3, 2, 3, 2, 3, 3, 2

Offset: 1

Lyle P. Blosser (lyleblosser(AT)att.net), May 18 2010

Sequence first term represents January 2000. Sequence repeats after 4800 terms, representing 400 years in the Gregorian calendar system.

Actual number of days in a month can be determined by adding 28 to the value of the sequence term representing the month in question.

			a(1) = 3 -> January 2000 has 31 days (3+28), a(2) = 1 -> February 2000 has 29 days (1+28), a(3) = 3 -> March 2000 has 31 days (3+28).

Lyle P. Blosser, Table of n, a(n) for n = 1..4800
Index entries for sequences related to calendars

Cf. A178054. If a(n) is the n-th term in A178054 and b(n) is the n-th term in A178055, then a(n) + b(n) (modulus 7) = a(n+1)

dys[{y_,m_,1}]:=If[m==12,DateDifference[{y,m,1},{y+1,1,1}],DateDifference[ {y,m,1},{y,m+1,1}]][[1]]; Mod[#,7]&/@(dys/@ Flatten[Table[{y,m,1},{y,2000,2010},{m,12}],1])  (* Harvey P. Dale, Sep 04 2020 *)

A178054 Numbers representing the index of the day of week for the first day of the month in the Gregorian calendar.

Original entry on oeis.org

6, 2, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5, 1, 4, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6, 2, 5, 5, 1, 3, 6, 1, 4, 0, 2, 5, 0, 3, 6, 6, 2, 4, 0, 2, 5, 1, 3, 6, 1, 4, 0, 1, 4, 6, 2, 4, 0, 3, 5, 1, 3, 6, 2, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4, 0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5, 1, 4, 4, 0, 2, 5, 0, 3, 6, 1, 4, 6, 2, 5, 6, 2, 4, 0, 2, 5, 1

Offset: 1

Lyle P. Blosser (lyleblosser(AT)att.net), May 18 2010

The index is 0-based, so 0 = Sunday, 1 = Monday, 2 = Tuesday, 3 = Wednesday, 4 = Thursday, 5 = Friday, 6 = Saturday.
The first term in the sequence represents the day of the week index for January 1, A.D. 2000.
The sequence repeats after 4800 terms, representing 400 years in the Gregorian calendar system.

			a(1) = 6, so day of week for January 1, 2000 is Saturday; a(2) = 2, so day of week for February 1, 2000 is Tuesday; a(3) = 3, so day of week for March 1, 2000 is Wednesday.

Arthur Benjamin and Michael Shermer, Secrets of Mental Math, First Edition, Three Rivers Press, 2006, p. 215.

Lyle P. Blosser, Table of n, a(n) for n = 1..4800
Index entries for sequences related to calendars

Cf. A178055.

a(n+1) = (a(n) + A178055(n)) (mod 7).

A160106 Decimal representation of Bernays's number, 67^257^729.

Original entry on oeis.org

3, 2, 5, 9, 1, 5, 3, 8, 4, 7, 9, 8, 6, 1, 8, 9, 9, 1, 6, 2, 5, 7, 1, 7, 8, 0, 6, 8, 8, 5, 0, 1, 5, 8, 8, 1, 1, 2, 0, 8, 6, 2, 1, 1, 5, 0, 5, 8, 7, 7, 8, 0, 7, 7, 6, 4, 5, 0, 7, 2, 8, 7, 0, 2, 5, 0, 4, 8, 0, 9, 7, 0, 2, 1, 2, 5, 0, 1, 7, 4, 1, 3, 3, 1, 7, 5, 6, 1, 7, 3, 6, 5, 3, 6, 4, 5, 8, 7, 4, 3, 2, 3, 5, 3, 7

Offset: 1

Lewis Mammel (l_mammel(AT)att.net), May 02 2009

This is a finite sequence of length:
12669674935126608420432141630855015714031380133279087897111823021713
56811328908882531121111469241905999472837913948238279755189743349761
48523228801813277516107342082973093097725413191748277420852876334406
94876293314725026209146791804598489379530361645466750631479593491258
89808249942992766773762667299010546238077478887602330971928923721941
72430386014378023796026916142427291438343856787929901324301858848058
27529209171651712159083473169942927988975800558560613650527749524532
75191774503837456493065661204570029626133921181521620538048404123145
80067317493063106206968226133732232940295274157977766381479776103292
42109055590354062378067741707188662030279086463891262509315257332626
67660622430734107904225269523607105245934662799643886003767606189798
57787550338482005946448422968364423839287672830452222083405995953816
36203273931424615452013250308765219156613666060842449019621385654602
20267721814801560898692089207050744121863093763466729360829090007340
19845582687744823456294708029903891488031593815746468873765082722973
55869028659436217274868023452405819990037705937486501551418694155825
46884222339479672918917024200948456377272821591381189093132349850355
90405444255979897423051268599606922301116055394691960425916429039897
40352095868171539874185632233360706548132174778016724460684684331817
19808367766356367096522727921316089451547342396490948067779940625178
88020116160602011047647958441543061184800996681742861951458927608369
31921303463436907590593465227992579980690076538677526802642563241223
58778978568413308707865022089920596975426734290393003094530833538477
51070253043054985292670186337562849238518822912544387932065661784941
62666108221075583052234535354001732258294144569659354587932951541940
4998441803274619168045467726087340720754974495397486708986
... which is 1 greater than the abscissa of the common logarithm of the number.
From Daniel Forgues, Jan 14 2012: (Start)
The length 1.2669... * 10^1757 (shown above) itself has 1758 digits!
The last 100 digits of Bernays's number are 67^257^729 mod 10^100 =
99736332723695669681470601458905407678415512345606116119671058321958
37693587043524719438498607119427. (End)

Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (first 2223 digits from Lewis Mammel)
P. Bernays, Sur le platonisme dans les mathématiques, L'Enseignement Mathématique 34 (1935), pp. 52-69. (See p. 61.)
P. Bernays, Platonism in mathematics, translated from the French by C. D. Parsons.
Efunda Engineering Fundamentals, Series Expansion of Exponential and Logarithmic Functions
K. Podnieks, Towards a real Finitism?

nbrdgt = 105; f[base_, exp_] := RealDigits[ 10^FractionalPart[ N[exp*Log10[base], nbrdgt + Floor[ Log10[ exp]] + 2]], 10, nbrdgt][[1]]; f[67, 257^729] (* and the last 100 digits computed by PowerMod[67, 257^729, 10^100] *) (* Robert G. Wilson v, Aug 09 2016 *)

/* bc script for cygwin bash shell or other UNIX environment */
/* Explicitly scale to 4000 decimal places */
i=10^4000
/* Calculate natural log of 100/67 */
b=100*i/67
c=i*(b-i)/(b+i)
x=c
s=0
for( j=1 ; x/j >0 ; j=j+2 ){
s = s + x/j
x = x*c/i*c/i
j /* progress mark */
}
s=2*s
/* Now s is the natural log of 100/67 */
/* Calculate natural log of sqrt(sqrt(10)) */
b=sqrt(10*i*i)
b=sqrt(b*i)
c=i*(b-i)/(b+i)
x=c
t=0
for( j=1 ; x/j >0 ; j=j+2 ){
t = t + x/j
x = x*c/i*c/i
j /* progress mark */
}
t=2*t
/* Now t is the integer part of 10^4000 * ln sqrt(sqrt(10)) */
ln10=4*t
/* ln10 is the integer part of 10^4000 * ln 10 */
ln67 = 2*ln10 - s
/* ln67 is the integer part of 10^4000 * ln 67 */
lg67 = ln67*i/ln10
/* lg67 is the integer part of 10^4000 * log_10 67 */
a=257^729
lgb = a*lg67
/* lgb is the integer part of 10^4000 * log_10 67^257^729 */
absc = lgb/i
/* absc is the abscissa of lgb, and its value is one less than the
number of decimal digits in Bernays's number */
mant = lgb - i*absc
/* Find number of digits in abscissa */
x=absc
for( nab=0 ; x>0 ; nab++ ) x = x/10
/* reduce the scale by nab */
mant = mant/10^nab
ln10 = ln10/10^nab
i = i/10^nab
/* find ln 10^mant */
lnmant = mant*ln10/i
/* calculate exp(lnmant) to get leading digits of Bernays's number */
fac=1
x=i
n=0
for( j=0 ; x/fac > 0 ; j++ ){
n = n + x/fac
x=x*lnmant/i
fac = fac*(j+1)
j /* progress mark */
}
/* display abscissa of log_10 ( Bernays's number ) */
absc
/* Display leading digits of Bernays's number.
( Truncation is to avoid displaying round-off error )*/
n / 10^20

Bernays's number is 67^257^729. The length and values of the sequence of its decimal representation is found by calculating its common logarithm by the formula, 257^729 * log_10(67) using an extended precision of 4000 digits. The number of digits of Bernays's number is given by the abscissa plus one, and the initial sequence is calculated from exponentiation of the mantissa.

A165968 Number of pairings disjoint to a given pairing, and containing a given pair not in the given pairing.

Original entry on oeis.org

0, 1, 2, 10, 68, 604, 6584, 85048, 1269680, 21505552, 407414816, 8535396256, 195927013952, 4890027052480, 131842951699328, 3818743350945664, 118253903175951104, 3898687202158805248, 136339489775029813760, 5040776996774472673792

Offset: 1

Lewis Mammel (l_mammel(AT)att.net), Oct 02 2009

nonn

The formula is derived by an application of the principle of inclusion and exclusion.
In reference to A053871, it is observed that the set of pairings disjoint to a given pairing can be partitioned into 2n-2 equivalent sets according to the 2n-2 pairs containing a given item. So it is seen that each term of that sequence must be divisible by 2n-2, giving the corresponding term of this sequence. However, the formula given here is derived independently.
Hankel transform of a(n+1) is A168467. Binomial transform of a(n+1) is A001147(n+1). - Paul Barry, Jan 26 2011
a(n) is a subset of the set of pairings of the first 2n integers (A001147) in another way: forbidding pairs of the form (2k,2k+1) for all k. - Olivier Gérard, Feb 08 2011

			a(1) = 0 trivially.
a(2) = 1 since there is a unique pairing disjoint to the canonical pairing, 01 23, and containing any of the 4 pairs not in the canonical pairing.
a(3) = 2 since there are 2 pairings disjoint to the canonical pairing, 01 23 45, and containing the pair 02, not in the canonical pairing: 02 14 35 and 02 15 34.

John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, Chapter 3

Michael De Vlieger, Table of n, a(n) for n = 1..405
Jean-Luc Baril, Avoiding patterns in irreducible permutations, Discrete Mathematics and Theoretical Computer Science, Vol 17, No 3 (2016).
Marilena Barnabei, Niccolò Castronuovo, and Matteo Silimbani, Hertzsprung patterns on involutions, arXiv:2412.03449 [math.CO], 2024. See p. 10.
Célia Biane, Greg Hampikian, Sergey Kirgizov, and Khaydar Nurligareev, Endhered patterns in matchings and RNA, arXiv:2404.18802 [math.CO], 2024. See pp. 8-9.

Cf. A001147, the double factorial.
Cf. also A053871.
Cf. A168467.

a:= n-> add((-1)^(n-k-1)*binomial(n-1,k)*(2*(k+1))!/(2^(k+1)*(k+1)!), k=0..n-1):
seq(a(n), n=0..20);

a[n_] := Sum[(-1)^(n-k-2)* Binomial[n-2, k]*(2*(k+1))!/(2^(k+1)*(k+1)!) , {k, 0, n-2}]; a /@ Range[20]
(* Jean-François Alcover, Jul 11 2011, after Maple *)
CoefficientList[Series[-1+1/(E^x*Sqrt[1-2*x]) + Sqrt[2]*DawsonF[1/Sqrt[2]] + Sqrt[-Pi/(2*E)]*Erf[Sqrt[-1/2+x]],{x,0,20}],x]*Range[0,20]! (* Vaclav Kotesovec, Feb 04 2014 *)

define a(n)
{
    auto sign, i,s;
    s=0; sign = 1;
    for ( i=0 ; i<=n-1 ; i++ ) {
        s = s + sign * ffac(n-1-i) * c( n-2, i );
        sign = sign * -1;
    }
    return s;
}
/* returns (2n-1)!! */
define ffac( n )
{
    if ( n <= 1 ) return 1;
    return  (2*n-1)* ffac(n-1);
}
/* returns combinations of n things taken i at a time */
define c(n,i)
{
    auto j,s;
    s=1;
    if ( n < 0 ) return 0;
    for ( j=0 ; j

Formula

a(n) = (2n-3)!! - C(n-2,1) * (2n-5)!! + ... +/- C(n-2,n-1)*3!! -/+ 1.

a(n) = (2*n-4)*a(n-1) +(2*n-6)*a(n-2) for n>2. - Gary Detlefs, Jul 11 2010

G.f.: x/(1-2x/(1-3x/(1+x-4x/(1-5x/(1+x-6x/(1-7x/(1+x-8x/(1-... (continued fraction). - Paul Barry, Jan 26 2011

a(n) = Sum_{k=0..n-1} (-1)^(n-k-1)*C(n-1,k)*(2*(k+1))!/(2^(k+1)*(k+1)!). - Paul Barry, Jan 26 2011

Conjecture: a(n) +2*(-n+1)*a(n-1) +2*(-n+2)*a(n-2)=0. - R. J. Mathar, Nov 15 2012

G.f.: x/G(0) where G(k) = 1 - 2*x*(2*k+1) - x^2*(2*k+2)*(2*k+3)/G(k+1) (continued fraction). - Sergei N. Gladkovskii, Jan 13 2013

G.f.: Q(0)-1, where Q(k) = 1 - x*(k+1)/( x*(k+1) - (1 +x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 21 2013

a(n) ~ 2^(n-1/2) * n^(n-1) / exp(n+1/2). - Vaclav Kotesovec, Feb 04 2014

a(n) = A053871(n)/(2n-2) for n>1.

a(n) * (2n-2) satisfies the recurrence of A053871, so Detlef's conjecture was correct. And if we rewrite Mathar's conjecture as b(n) = 2*(n-1)*b(n-1) +2*(n-2)*b(n-2) it becomes quite clear that Mathar's b(n) = a(n-1). - Sergey Kirgizov, Jun 03 2022

A167051 Start at 1, then add the first term (which is one here) plus 1 for the second term; then add the second term plus 2 for the third term; then add the third term to the sum of the first and second term; this gives the fourth term. Restart the sequence by adding 1 to the fourth term, etc. (From a sixth grade math extra credit assignment).

Original entry on oeis.org

1, 2, 4, 7, 8, 10, 25, 26, 28, 79, 80, 82, 241, 242, 244, 727, 728, 730, 2185, 2186, 2188, 6559, 6560, 6562, 19681, 19682, 19684, 59047, 59048, 59050, 177145, 177146, 177148, 531439, 531440, 531442, 1594321, 1594322, 1594324, 4782967, 4782968, 4782970, 14348905

Offset: 1

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Author

Chris Rice (cwrice(AT)research.att.com), Oct 27 2009

Keywords

nonn

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1000

Programs

PARI

seq(n)={my(a=vector(n)); a[1]=1; for(n=2, #a, my(t=n%3); a[n]=a[n-1]+if(t==2, 1, if(t==0, 2, a[n-2]+a[n-3]))); a} \\ Andrew Howroyd, Apr 13 2021

PARI

Vec((1 + 2*x + 4*x^2 + 3*x^3 - 6*x^5)/((1 - x)*(1 + x + x^2)*(1 - 3*x^3)) + O(x^40)) \\ Andrew Howroyd, Apr 13 2021

Formula

a(n) = a(n-1) + 1 for n mod 3 == 2;

a(n) = a(n-1) + 2 for n mod 3 == 0;

a(n) = a(n-1) + a(n-2) + a(n-3) for n mod 3 == 1 and n > 1.

G.f.: x*(1 + 2*x + 4*x^2 + 3*x^3 - 6*x^5)/((1 - x)*(1 + x + x^2)*(1 - 3*x^3)). - Andrew Howroyd, Apr 13 2021

A147653 Number of letters (including spaces) in last name of n-th President of U.S.A.

Original entry on oeis.org

10, 5, 9, 7, 6, 5, 7, 9, 8, 5, 4, 6, 8, 6, 8, 7, 7, 5, 5, 8, 6, 9, 8, 9, 8, 9, 4, 6, 7, 8, 6, 9, 6, 10, 7, 7, 5, 4, 6, 6, 4, 7, 4, 5, 5, 5

Offset: 1

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Author

Jay Jacob Wind (jay.wind(AT)att.net), Nov 09 2008

Keywords

easy
nonn

Comments

a(8) = 9 because "Van Buren" has 9 letters, including the space.

Links

Wikipedia, Presidents of the United States

Crossrefs

Cf. A008745, A072550, A084989, A090232.

Extensions

a(45) from Jon E. Schoenfield, Jan 20 2017

a(46) from Jon E. Schoenfield, Jan 20 2021

A145383 Indices in A050791 for the subsequence A141326.

Original entry on oeis.org

1, 3, 6, 10, 13, 17, 20, 23, 30, 37, 38, 42, 44, 48, 51, 55, 58, 63, 68, 72, 73, 77, 83, 84, 87, 90, 94, 97, 99, 102, 106, 108, 111, 115, 121, 123, 125, 130, 133, 134, 136, 140, 142, 146, 150, 153, 156, 159, 164, 167, 169, 172

Offset: 1

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Author

Lewis Mammel (l_mammel(AT)att.net), Oct 10 2008

Keywords

easy
nonn

Comments

The first difference of this sequence is used to define the sequence A145384, which is of particular interest.

Links

Lewis Mammel, Table of n, a(n) for n = 1..122

Crossrefs

Cf. A050791, values of z where 1+z^3 = x^3 + y^3.

Cf. A141326, a subsequence of A050791 defined by a simple formula.

Cf. A145384, a sequence based on the first difference of this sequence.

Formula

The defining equation is A050791(a(n)) = A141326(n).

A141326 Subsequence of 'Fermat near misses' which is generated by a simple formula based on the cubic binomial expansion along with formulas for the corresponding terms in the expression, x^3 + y^3 = z^3 + 1.

Original entry on oeis.org

12, 150, 738, 2316, 5640, 11682, 21630, 36888, 59076, 90030, 131802, 186660, 257088, 345786, 455670, 589872, 751740, 944838, 1172946, 1440060, 1750392, 2108370, 2518638, 2986056, 3515700, 4112862, 4783050, 5531988, 6365616, 7290090, 8311782, 9437280, 10673388

Offset: 1

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Author

Lewis Mammel (l_mammel(AT)att.net), Aug 03 2008

Keywords

easy
nonn

Comments

From Lewis Mammel (l_mammel(AT)att.net), Aug 21 2008: (Start)

In Ramanujan's parametric equation: (ax+y)^3 + (b+x^2y)^3 = (bx+y)^3 + (a+x^2y)^3

where a^2 + ab + b^2 = 3xy^2.

This sequence is obtained by setting a=0, y=1 and finding the solution to b^2=3x:

b=3n, x=3n^2. (End)

Examples

For a(1)=12: 1 + 12^3 = 9^3 + 10^3 = 1729.

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Tito Piezas III and Eric Weisstein, Diophantine Equation--3rd Powers [Lewis Mammel (l_mammel(AT)att.net), Aug 21 2008]

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A050791, A050792, A050793, A050794.

Programs

PARI

Vec(6*x*(2 + 15*x + 18*x^2 + x^3) / (1 - x)^5 + O(x^40)) \\ Colin Barker, Oct 26 2019

Formula

a(n) = 9*n^4 + 3*n, with b(n) = 9*n^4 and c(n) = 9*n^3 + 1 we have 1 + a(n)^3 = b(n)^3 + c(n)^3.

From Colin Barker, Oct 25 2019: (Start)

G.f.: 6*x*(2 + 15*x + 18*x^2 + x^3) / (1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.

a(n) = 3*(n + 3*n^4).

(End)

Extensions

Edited by Joerg Arndt, Oct 26 2019

A143936 Subsequence of A050791, "Fermat near misses", generated by iteration of a linear form derived from Ramanujan's parametric formula for equal sums of two pairs of cubes.

Original entry on oeis.org

5262, 2262756, 972979926, 418379105532, 179902042398942, 77357459852439636

Offset: 1

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Author

Lewis Mammel (l_mammel(AT)att.net), Sep 05 2008

Keywords

nonn

Comments

The formulas give an approximately geometric progression of values, z, such that 1 + z^3 = x^3 + y^3, along with the values for x and y. Iteration yields large values of x,y and z presumably unobtainable by exhaustive search.

Examples

1 + 5262^3 = 4528^3 + 3753^3 = 145697644729 1 + 2262756^3 = 1947250^3 + 1613673^3 = 11585457155467377217 1 + 972979926^3 = 837313192^3 + 693875529^3 = 921110304262410135315034777

References

Charles Edward Sandifer, The Early Mathematics of Leonhard Euler, 2007, pp. 102-103.

Links

Wolfram Mathworld, Diophantine Equation 3rd Powers

Crossrefs

Cf. A050791, A141326.

Programs

Other

/* File: form.bc Usage: bc form.bc ( In UNIX shell, e.g. bash on Cygwin ) */ define a(x){ return( 321*x^2 + 216*x + 36 ); } define b(x){ return( sqrt(a(x)) ); } define n(z){ auto a,x; x=3; a = 215*z+12*b(z)+72 ; a;b(a); return(v(a)); } define v(z){ auto a,b,x,y,i,j,k,l; a = z; b = ( a + b(a) )/2; a = -a; x=3; y = 1-a*x; i=a*x+y; j=b+x^2*y; k=b*x+y; l=a+x^2*y; -a; b; i;j;k;l; i^3+j^3; k^3+l^3; return ( -a ); } z=144; v(z) ; z=n(z); z=n(z); z=n(z); /* ... etc. */

Formula

In Ramanujan's parametric formula:

(a*x+y)^3 + (b+x^2*y)^3 = (b*x+y)^3 + (a+x^2*y)^3

with

a^2 + a*b + b^2 = x*y^2,

we set x=3, ax+y=1 and obtain a quadratic equation for b in terms of a

( Since 'a' is always negative we write it explicitly as '-a' and solve for positive 'a' )

The surd of the quadratic formula then becomes:

sqrt(321*a^2 + 216*a + 36)

and we require that this be an integer. After finding an initial value of 'a' which satisfies this condition by inspection of the sequence A050791, we use Euler's method to find the bilinear recursion: ( with s_i == sqrt(321*a_i^2 + 216*a_i + 36) )

a_i+1 = 215*a_i + 12*s_i + 72

s_i+1 = 215*s_i + 3852*a_i + 1296

and these yield the values of x,y and z from Ramanujan's formula.

A145384 The number of terms of A050791 bracketed by successive terms of A141326.

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 2, 2, 6, 6, 0, 3, 1, 3, 2, 3, 2, 4, 4, 3, 0, 3, 5, 0, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 5, 1, 1, 4, 2, 0, 1, 3, 1, 3, 3, 2, 2, 2, 4, 2, 1, 2, 4, 2, 0, 1, 2, 3, 1, 1, 1, 3, 0, 3, 1, 0, 3, 1, 1, 4, 2, 2, 1, 3, 3, 1, 2, 0, 3, 2, 5, 1, 1, 3, 6, 2, 4, 1, 0, 5, 2, 2, 2, 2, 3, 2, 3, 3, 0, 1

Offset: 1

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Author

Lewis Mammel (l_mammel(AT)att.net), Oct 10 2008

Keywords

nonn

Comments

A141326 is a simply generated subsequence of A050791 and by observation it forms a natural measure of the parent sequence. The first several hundred terms of the parent sequence not belonging to A141326 are bracketed into groups with a small integral number of terms ( including 0 ) by the successive terms of the subsequence, A141326.

a(107),a(108) are the first occurrence of 2 consecutive 0's and a(119),a(120),a(121) are the first occurrence of 3 consecutive 0's. This leads to the following conjecture:

-> 0 as n ->inf

where = ( sum m=1,n of a(m) )/n

Examples

0 = number of terms of A050791 preceding the first term of A141326 1 = number of terms of A050791 between the first and 2nd terms of A141326 2 = number of terms of A050791 between the 2nd and 3rd terms of A141326

Links

Lewis Mammel, Table of n, a(n) for n = 1..122

Crossrefs

Cf. A145383, A141326, A050791

Formula

a(1) = A145383(1) - 1

a(n) = A145383(n) - A145383(n-1) - 1 ; n>1

cp's OEIS Frontend

User: åtte

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A145383 Indices in A050791 for the subsequence A141326.

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A141326 Subsequence of 'Fermat near misses' which is generated by a simple formula based on the cubic binomial expansion along with formulas for the corresponding terms in the expression, x^3 + y^3 = z^3 + 1.

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PARI